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Vol. 2, No. 3, 2006 ISSN 1556-6706

SCIENTIA MAGNA

Edited by Department of Mathematics Northwest University, P. R. China

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Vol. 2, No. 3, 2006 ISSN 1556-6706

SCIENTIA MAGNA

Edited by

Department of Mathematics Northwest University Xi’an, Shaanxi, P.R.China

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ii Contributing to Scientia Magna

Authors of papers in science (mathematics, physics, philosophy, psychology, sociology, linguistics) should submit manuscripts, by email, to the Editor-in-Chief:

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Dr. Huaning Liu, Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China. E-mail: [email protected]

Prof. Yuan Yi, Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China. E-mail: [email protected]

Dr. Zhefeng Xu, Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China. E-mail: [email protected]; [email protected]

Dr. Tianping Zhang, College of Mathematics and Information Science, Shaanxi Normal University, Xi’an, Shaanxi, P.R.China. E-mail: [email protected]

iii Contents

A. Majumdar : A note on the Pseudo-Smarandache function 1 S. Gupta : Primes in the Smarandache deconstructive sequence 26 S. Zhang and C. Chen : Recursion formulae for Riemann zeta function and Dirichlet series 31 A. Muktibodh : Smarandache semiquasi near-rings 41 J. S´andor: On exponentially harmonic numbers 44

T. Ja´ıy´eo.l´a: Parastrophic invariance of Smarandache quasigroups 48 M. Karama : Perfect powers in Smarandache n-expressions 54

T. Ja´ıyéo.lá: Palindromic permutations and generalized Smarandache palindromic permutations 65 J. Sándor: On certain inequalities involving the Smarandache function 78 A. Muktibodh : Sequences of pentagonal numbers 81 J. Gao : On the additive analogues of the simple function 88 W. Zhu : On the primitive numbers of power p 92 A. Vyawahare : Smarandache sums of products 96 N. Yuan : On the solutions of an equation involving the Smarandache dual function 104 H. Zhou : An infinite series involving the Smarandache power function SP (n) 109

iv Scientia Magna Vol. 2 (2006), No. 3, 1-25

A note on the Pseudo-Smarandache function

A.A.K. Majumdar1 Department of Mathematics, Jahangirnagar University, Savar, Dhaka 1342, Bangladesh

Abstract This paper gives some results and observations related to the Pseudo-Smarandache function Z(n). Some explicit expressions of Z(n) for some particular cases of n are also given. Keywords The Pseudo-Smarandache function, Smarandache perfect square, equivalent.

§1. Introduction

The Pseudo-Smarandache function Z(n), introduced by Kashihara [1], is as follows : Deﬁnition 1.1. For any integer n ≥ 1, Z(n) is the smallest positive integer m such that 1 + 2 + ··· + m is divisible by n. Thus, ½ ¾ m(m + 1) Z(n) = min m : m ∈ N: n | . (1.1) 2

As has been pointed out by Ibstedt [2], an equivalent deﬁnition of Z(n) is Deﬁnition 1.2. √ Z(n) = min{k : k ∈ N: 1 + 8kn is a perfect square}.

Kashihara [1] and Ibstedt [2] studied some of the properties satisfied by Z(n). Their findings are summarized in the following lemmas: Lemma 1.1. For any m ∈ N, Z(n) ≥ 1. Moreover, Z(n) = 1 if and only if n = 1, and Z(n) = 2 if and only if n = 3. Lemma 1.2. For any prime p ≥ 3, Z(p) = p − 1. Lemma 1.3. For any prime p ≥ 3 and any k ∈ N, Z(pk) = pk − 1. Lemma 1.4. For any k ∈ N, Z(2k) = 2k+1 − 1. Lemma 1.5. For any composite number n ≥ 4, Z(n) ≥ max{Z(N): N | n}. In this paper, we give some results related to the Pseudo-Smarandache function Z(n). In §2, we present the main results of this paper. Simple explicit expressions for Z(n) are available for particular cases of n. In Theorems 2.1 − 2.11, we give the expressions for Z(2p), Z(3p), Z(2p2), Z(3p3), Z(2pk), Z(3pk), Z(4p), Z(5p), Z(6p), Z(7p) and Z(11p), where p is a prime and k(≥ 3) is an integer. Ibstedt [2] gives an expression for Z(pq) where p and q are distinct primes. We give an alternative expressions for Z(pq), which is more efficient from the computational point of view. This is given in Theorem 2.12, whose proof shows that the solution of Z(pq) involves the solution of two Diophantine equations. Some particular cases of Theorem

1On Sabbatical leave from: Ritsumeikan Asia-Paciﬁc University, 1-1 Jumonjibaru, Beppu-shi, Oita-ken, Japan. 2 A.A.K. Majumdar No. 3

2.12 are given in Corollaries 2.1 − 2.16. We conclude this paper with some observations about the properties of Z(n), given in four Remarks in the last §3.

§2. Main Results

We ﬁrst state and prove the following results. k(k+1) Lemma 2.1. Let n = 2 for some k ∈ N. Then, Z(n) = k. Proof. Noting that k(k + 1) = m(m + 1) if and only if k = m, the result follows. The following lemma gives lower and upper bounds of Z(n). Lemma 2.2. 3 ≤ n ≤ 2n − 1 for all n ≥ 4. m(m+1) Proof. Letting f(m) = 2 , m ∈ N, see that f(m) is strictly increasing in m with f(2) = 3. Thus, Z(n) = 2 if and only if n = 3. This, together with Lemma 1.1, gives the lower bound of Z(n) for n ≥ 4. Again, since n | f(2n − 1), it follows that Z(n) cannot be greater than 2n − 1. Since Z(n) = 2n − 1 if n = 2k for some k ∈ N, it follows that the upper bound of Z(n) in Lemma 2.2 cannot be improved further. However, the lower bound of Z(n) can be improved. For example, since f(4) = 10, it follows that Z(n) ≥ 5 for all n ≥ 11. A better lower bound of Z(n) is given in Lemma 1.5 for the case when n is a composite number. In Theorems 2.1 − 2.4, we give expressions for Z(2p),Z(3p),Z(2p2) and Z(3p2) where p ≥ 5 is a prime. To prove the theorems, we need the following results. Lemma 2.3. Let p be a prime. Let an integer n(≥ p) be divisible by pk for some integer k(≥ 1). Then, pk does not divide n + 1 (and n − 1). Lemma 2.4. 6 | n(n + 1)(n + 2) for any n ∈ N. In particular, 6 | (p2 − 1) for any prime p ≥ 5. Proof. The ﬁrst part is a well-known result. In particular, for any prime p ≥ 5, 6 | (p − 1)p(p + 1). But since p(≥ 5) is not divisible by 6, it follows that 6 | (p − 1)(p + 1). Theorem 2.1. If p ≥ 5 is a prime, then   p − 1, if 4 | (p − 1); Z(2p) =  p, if 4 | (p + 1).

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(2p) = min m : 2p | = min m : p | . (1) 2 4

If p | m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3). Thus, the minimum m in (1) may be taken as p − 1 or p depending on whether p − 1 or p + 1 respectively is divisible by 4. We now consider the following two cases that may arise : Case 1 : p is of the form p=4a+1 for some integer a ≥ 1. In this case, 4 | (p − 1), and hence, Z(2p) = p − 1. Case 2 : p is of the form p = 4a + 3 for some integer a ≥ 1. Here, 4 | (p + 1) and hence, Z(2p) = p. Vol. 2 A note on the Pseudo-Smarandache function 3

Theorem 2.2. If p ≥ 5 is a prime, then   p − 1, if 3 | (p − 1); Z(3p) =  p, if 3 | (p + 1).

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(3p) = min m : 3p | = min m : p | . (2) 2 6

If p | m(m + 1), then p must divide either m or m+1, but not both (by Lemma 2.3). Thus, the minimum m in (2) may be taken as p − 1 or p according as p − 1 or p + 1 respectively is divisible by 6. But, since both p − 1 and p + 1 are divisible by 2, it follows that the minimum m in (2) may be taken as p − 1 or p according as p − 1 or p + 1 respectively is divisible by 3. We now consider the following two possible cases that may arise : Case 1 : p is of the form p = 3a + 1 for some integer a ≥ 1. In this case, 3 | (p − 1), and hence, Z(3p) = p − 1. Case 2 : p is of the form p = 3a + 2 for some integer a ≥ 1. Here, 3 | (p + 1), and hence, Z(3p) = p. Theorem 2.3. If p ≥ 3 is a prime, then Z(2p2) = p2 − 1. Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(2p2) = min m : 2p2 | = min m : p2 | . (3) 2 4

If p2|m(m + 1), then p2 must divide either m or m + 1, but not both (by Lemma 2.3). Thus, the minimum m in (3) may be taken as p2 − 1 if p2 − 1 is divisible by 4. But, since both p − 1 and p + 1 are divisible by 2, it follows that 4 | (p − 1)(p + 1). Hence, Z(2p2) = p2 − 1. Theorem 2.4. If p ≥ 5 is a prime, then Z(3p2) = p2 − 1. Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(3p2) = min m : 3p2 | = min m : p2 | . (4) 2 6

If p2|m(m + 1), then p2 must divide either m or m + 1, but not both (by Lemma 2.3). Thus, the minimum m in (4) may be taken as p21 if p2 − 1 is divisible by 6. By Lemma 2.4, 6 | (p2 − 1). Consequently, Z(2p2) = p2 − 1.

Deﬁnition 2.1. A function g :N → N is called multiplicative if and only if g(n1n2) =

g(n1)g(n2) for all n1, n2 ∈ N with (n1, n2) = 1. Remark 2.1. From Lemma 1.2 and Theorem 2.1, we see that Z(2p) 6= 3(p−1) = Z(2)Z(p) for any odd prime p. Moreover, Z(3p2) = p2 − 1 6= Z(2p2) + Z(p2). These show that Z(n) is neither additive nor multiplicative, as has already been noted by Kashihara [1]. The expressions for Z(2pk) and Z(3pk) for k ≥ 3 are given in Theorem 2.5 and Theorem 2.6 respectively. For the proofs, we need the following results: Lemma 2.5. (1) 4 divides 32k − 1 for any integer k ≥ 1. (2) 4 divides 32k+1 + 1 for any integer k ≥ 0. 4 A.A.K. Majumdar No. 3

Proof. (1) Writing 32k − 1 = (3k − 1)(3k + 1), the result follows immediately. (2) The proof is by induction on k. The result is clearly true for k = 0. So, we assume that the result is true for some integer k, so that 4 divides 32k+1 + 1 for some k. Now, since 32k+3 + 1 = 9(32k+1 + 1) − 8, it follows that 4 divides 32k+3 + 1, completing the induction. Lemma 2.6. (1) 3 divides 22k − 1 for any integer k ≥ 1. (2) 3 divides 22k+1 + 1 for any integer k ≥ 0. Proof. (1) By Lemma 2.4, 3 divides (2k − 1)2k(2k + 1). Since 3 does not divide 2k, 3 must divide (2k − 1)(2k + 1) = 22k − 1. (2) The result is clearly true for k = 0. To prove by induction, the induction hypothesis is that 3 divides 22k+1 + 1 for some k. Now, since 22k+3 + 1 = 4(32k+1 + 1) − 3, it follows that 3 divides 22k+3 + 1, so that the result is true for k + 1 as well, completing the induction. Theorem 2.5. If p ≥ 3 is a prime and k ≥ 3 is an integer, then   pk, if 4 | (p − 1) and k is odd; Z(2pk) =  pk − 1, otherwise.

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(2pk) = min m : 2pk | = min m : pk | . (5) 2 4

If pk|m(m+1), then pk must divide either m or m+1, but not both (by Lemma 2.3). Thus, the minimum m in (5) may be taken as pk − 1 or pk according as pk − 1 or pk is respectively divisible by 4. We now consider the following two possibilities: Case 1 : p is of the form 4a + 1 for some integer a ≥ 1. In this case, pk = (4a + 1)k = k 1 k−1 k−1 k (4a) + Ck (4a) + ··· + Ck (4a) + 1, showing that 4 | (p − 1). Hence, in this case, Z(2pk) = pk − 1. Case 2 : p is of the form 4a + 3 for some integer a ≥ 1. Here, pk = (4a + 3)k = k 1 k−1 k−1 k−1 k (4a) + Ck (4a) 3 + ··· + Ck (4a)3 + 3 . (1) If k ≥ 2 is even, then by Lemma 2.5, 4 | (3k − 1), so that 4 | (pk − 1). Thus, Z(2pk) = pk − 1. (2) If k ≥ 3 is odd, then by Lemma 2.5, 4 | (3k +1), and so 4 | (pk +1). Hence, Z(2pk) = pk. All these complete the proof of the theorem. Theorem 2.6. If p ≥ 3 is a prime and k ≥ 3 is an integer, then   pk, if 3 | (p + 1) and k is odd; Z(3pk) =  pk − 1, otherwise.

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(3pk) = min m : 3pk | = min m : pk | . (6) 2 6 Vol. 2 A note on the Pseudo-Smarandache function 5

If pk|m(m+1), then pk must divide either m or m+1, but not both (by Lemma 2.3). Thus, the minimum m in (6) may be taken as pk − 1 or pk according as pk − 1 or pk is respectively divisible by 6. We now consider the following two possible cases: Case 1 : p is of the form 3a + 1 for some integer a ≥ 1. In this case, pk = (3a + 1)k = k 1 k−1 k−1 k (3a) + Ck (3a) + ··· + Ck (3a) + 1, it follows that 3 | (p − 1). Thus, in this case, Z(3pk) = pk − 1. Case 2 : p is of the form 3a + 2 for some integer a ≥ 1. Here, pk = (3a + 2)k = k 1 k−1 k−1 k−1 k (3a) + Ck (3a) (2) + ··· + Ck (3a)2 + 2 . (1) If k ≥ 2 is even, then by Lemma 2.6, 3 | (2k − 1), so that 3 | (pk − 1). Thus, Z(3pk) = pk − 1. (2) If k ≥ 3 is odd, then by Lemma 2.6, 3 | (2k +1), and so 3 | (pk +1). Thus, Z(3pk) = pk. In Theorem 2.7–Theorem 2.9, we give the expressions for Z(4p), Z(5p) and Z(6p) respectively, where p is a prime. Note that, each case involves 4 possibilities. Theorem 2.7. If p ≥ 5 is a prime, then    p − 1, if 8 | (p − 1);   p, if 8 | (p + 1); Z(4p) =  3p − 1, if 8 | (3p − 1);   3p, if 8 | (3p + 1).

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(4p) = min m : 4p | = min m : p | . (7) 2 8

If p|m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3), and then 8 must divide either p − 1 or p + 1, In the particular case when 8 divides p − 1 or p + 1, the minimum m in (7) may be taken as p − 1 or p + 1 respectively. We now consider the following four cases may arise: Case 1 : p is of the form p = 8a + 1 for some integer a ≥ 1. In this case, 8 | (p − 1), and hence Z(4p) = p − 1. Case 2 : p is of the form p = 8a + 7 for some integer a ≥ 1. Here, 8 | (p + 1), and hence Z(4p) = p. Case 3 : p is of the form p = 8a + 3 for some integer a ≥ 1. In this case, 8 | (3p − 1), and hence Z(4p) = 3p − 1. Case 4 : p is of the form p = 8a + 5 for some integer a ≥ 1. Here, 8 | (3p + 1), and hence Z(4p) = 3p. Theorem 2.8. If p ≥ 7 is a prime, then    p − 1, if 10 | (p − 1);   p, if 10 | (p + 1); Z(5p) =  2p − 1, if 5 | (2p − 1);   2p, if 5 | (2p + 1). 6 A.A.K. Majumdar No. 3

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(5p) = min m : 5p | = min m : p | . (8) 2 10

If p|m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3), and then 5 must divide either m − 1 or m + 1, In the particular case when 5 divides p − 1 or p + 1, the minimum m in (8) may be taken as p − 1 or p + 1 respectively. We now consider the four below that may arise: Case 1 : p is a prime whose last digit is 1. In this case, 10 | (p−1), and hence Z(5p) = p−1. Case 2 : p is a prime whose last digit is 9. In such a case, 10 | (p + 1), and so Z(5p) = p. Case 3 : p is a prime whose last digit is 3. In this case, 5 | (2p − 1). Thus, the minimum m in (9) may be taken as 2p − 1. Hence Z(5p) = 2p − 1. Case 4 : p is a prime whose last digit is 7. Here, 5 | (2p + 1), and hence Z(5p) = 2p. Theorem 2.9. If p ≥ 5 is a prime, then    p − 1, if 12 | (p − 1);   p, if 12 | (p + 1); Z(6p) =  2p − 1, if 4 | (3p + 1);   2p, if 4 | (3p − 1).

Proof. ½ ¾ ½ ¾ m(m + 1) m(m + 1) Z(6p) = min m : 6p | = min m : p | . (9) 2 12

If p|m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3), and then 12 must divide either m − 1 or m + 1, In the particular case when 12 divides p − 1 or p + 1, the minimum m in (9) may be taken as p − 1 or p respectively. We now consider the four cases that may arise: Case 1 : p is of the form p = 12a + 1 for some integer a ≥ 1. In this case, 12 | (p − 1), and hence Z(6p) = p − 1. Case 2 : p is of the form p = 12a + 11 for some integer a ≥ 1. Here, 12 | (p + 1), and hence Z(6p) = p. Case 3 : p is of the form p = 12a + 5 for some integer a ≥ 1. In this case, 4 | (3p + 1). The minimum m in (10) may be taken as 3p, and hence Z(6p) = 3p. Case 4 : p is of the form p = 12a + 7 for some integer a ≥ 1. Here, 4 | (3p − 1), and hence Z(6p) = 3p − 1. It is possible to ﬁnd explicit expressions for Z(7p) or Z(11p), where p is a prime, as are given in Theorem 2.10 and Theorem 2.11 respectively, but it becomes more complicated. For example, in ﬁnding the expression for Z(7p), we have to consider all the six possibilities, while the expression for Z(11p) involves 10 alternatives. Vol. 2 A note on the Pseudo-Smarandache function 7

Proof: m(m + 1) m(m + 1) Z(7p) = min{m : 7p| } = min{m : p| }. (10) 2 14 If p|m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3), and then 7 must divide either m + 1 or m respectively. In the particular case when 12 divides p − 1 or p + 1, the minimum m in (10) may be taken as p − 1 or p respectively. We now consider the following six cases that may arise: Case 1 : p is of the form p = 7a+1 for some integer a ≥ 1. In this case, 7|(p−1). Therefore, Z(7p) = p − 1. Case 2 : p is of the form p = 7a + 6 for some integer a ≥ 1. Here, 7|(p + 1), and so, Z(7p) = p. Case 3 : p is of the form p = 7a + 2 for some integer a ≥ 1, so that 7|(3p + 1). In this case, the minimum m in (11) may be taken as 3p. That is, Z(7p) = 3p. Case 4 : p is of the form p = 7a + 5 for some integer a ≥ 1. Here, 7|(3p − 1), and hence, Z(7p) = 3p − 1. Case 5 : p is of the form p = 7a + 3 for some integer a ≥ 1. In this case, 7|(2p + 1), and hence, Z(7p) = 2p. Case 6 : p is of the form p = 7a + 4 for some integer a ≥ 1. Here, 7|(2p − 1), and hence, Z(7p) = 2p − 1. Theorem 2.11. For any prime p ≥ 13,   p − 1, if 11 | (p − 1);    p, if 11 | (p + 1);    2p − 1, if 11 | (2p − 1);    2p, if 11 | (2p + 1);   3p − 1, if 11 | (3p − 1); Z(7p) =   3p, if 11 | (3p + 1);    4p − 1, if 11 | (4p − 1);    4p, if 11 | (4p + 1);   5p − 1, if 11 | (5p − 1);   5p, if 11 | (5p + 1). , 8 A.A.K. Majumdar No. 3

Proof: m(m + 1) m(m + 1) Z(11p) = min{m : 11p| } = min{m : p| }. (11) 2 22 If p|m(m + 1), then p must divide either m or m + 1, but not both (by Lemma 2.3), and then 11 must divide either m + 1 or m respectively. In the particular case when 11 divides p − 1 or p + 1, the minimum m in (11) may be taken as p − 1 or p respectively. We have to consider the ten possible cases that may arise : Case 1 : p is of the form p = 11a + 1 for some integer a ≥ 1. In this case, 11|(p − 1), and so, Z(11p) = p − 1. Case 2 : p is of the form p = 11a + 10 for some integer a ≥ 1. Here, 11|(p + 1), and hence, Z(11p) = p. Case 3 : p is of the form p = 11a + 2 for some integer a ≥ 1. In this case, 11|(5p + 1), and hence, Z(11p) = 5p. Case 4 : p is of the form p = 11a + 9 for some integer a ≥ 1. Here, 11|(5p − 1), and hence, Z(11p) = 5p − 1. Case 5 : p is of the form p = 11a + 3 for some integer a ≥ 1. In this case, 11|(4p − 1), and hence, Z(11p) = 4p − 1. Case 6 : p is of the form p = 11a + 8 for some integer a ≥ 1. Here, 11|(4p + 1), and hence, Z(11p) = 4p. Case 7 : p is of the form p = 11a + 4 for some integer a ≥ 1. In this case, 11|(3p − 1), and hence, Z(11p) = 3p − 1. Case 8 : p is of the form p = 11a + 7 for some integer a ≥ 1. Here, 11|(3p + 1), and hence, Z(11p) = 3p. Case 9 : p is of the form p = 11a + 5 for some integer a ≥ 1. In this case, 11|(2p + 1), and hence, Z(11p) = 2p. Case 10 : p is of the form p = 11a + 6 for some integer a ≥ 1. Here, 11|(2p − 1), and hence, Z(11p) = 2p − 1. In Theorem 2.12, we give an expression for Z(pq), where p and q are two distinct primes. In this connection, we state the following lemma. The proof of the lemma is similar to, for example, Theorem 12.2 of Gioia [3], and is omitted here. Lemma 2.7. Let p and q be two distinct primes. Then, the Diophantine equation

qy − px = 1

has an inﬁnite number of solutions. Moreover, if (x0, y0) is a solution of the Diophantine equation, then any solution is of the form

x = x0 + qt, y = y0 + pt,

where t ≥ 0 is an integer. Theorem 2.12. Let p and q be two primes with q > p ≥ 5. Then,

Z(pq) = min{qy0 − 1, px0 − 1}, Vol. 2 A note on the Pseudo-Smarandache function 9

where

y0 = min{y : x, y ∈ N, qy − px = 1},

x0 = min{x : x, y ∈ N, px − qy = 1}.

Proof: Since m(m + 1) Z(pq) = min{m : pq| }, (12) 2 it follows that we have to consider the three cases below that may arise : Case 1 : When p|m and q|(m + 1). In this case, m = px for some integer x ≥ 1, m + 1 = qy for some integer y ≥ 1. From these two equations, we get the Diophantine equation

qy − px = 1.

By Lemma 2.7, the above Diophantine equation has inﬁnite number of solutions. Let

y0 = min{y : x, y ∈ N, qy − px = 1}.

For this y0, the corresponding x0 is given by the equation q0y − p0x = 1. Note that y0 and x0 cannot be both odd or both even. Then, the minimum m in (12) is given by

m + 1 = qy0 ⇒ m = qy0 − 1.

Case 2 : When p|(m + 1) and q|m. Here, m + 1 = px for some integer x ≥ 1, m = qy for some integer y ≥ 1. These two equations lead to the Diophantine equation. px − qy = 1. Let

x0 = min{x : x, y ∈ N, px − qy = 1}.

For this x0, the corresponding y0 is given by y0 = (px0 −1)/q. Here also, x0 and y0 both cannot be odd or even simultaneously. The minimum m in (12) is given by

m + 1 = px0 ⇒ m = px0 − 1.

Case 3 : When pq|(m+1). In this case, m = pq −1. But then, by Case 1 and Case 2 above, this does not give the minimum m. Thus, this case cannot occur. The proof of the theorem now follows by virtue of Case 1 and Case 2. Remark 2.2. Let p and q be two primes with q ≥ p ≥ 5. Let q = kp + ` for some integers k and ` with k ≥ 1 and 1 ≤ ` ≤ p − 1. We now consider the two cases given in Theorem 2.12 : Case 1 : When p|m and q|(m + 1). In this case, m = px for some integer x ≥ 1, m + 1 = qy = (kp + `)y for some integer y ≥ 1. From these two equations, we get

`y − (x − ky)p = 1 (2.1).

Case 2 : When p|(m + 1) and q|m. Here, m + 1 = px for some integer x ≥ 1, m = (kp + `)y for some integer y ≥ 1. These two equations lead to

(x − ky)p − `y = 1 (2.2). 10 A.A.K. Majumdar No. 3

In some particular cases, explicit expressions of Z(pq) may be found. These are given in the following corollaries. Corollary 2.1. Let p and q be two primes with q > p ≥ 5. Let q = kp+1 for some integer k ≥ 2. Then, Z(pq) = q − 1. Proof. From (2.1) with ` = 1, we get y − (x − ky)p = 1, the minimum solution of which is y = 1, x = ky = k. Then, the minimum m in (12) is given by

m + 1 = qy = q ⇒ m = q − 1.

Note that, from (2.2) with ` = 1, we have (x − ky)p − y = 1, with the least possible solution y = p − 1 (and x − ky = 1). Corollary 2.2. Let p and q be two primes with q > p ≥ 5. Let q = (k + 1)p − 1 for some integer k ≥ 1. Then, Z(pq) = q. Proof. From (2.2) with ` = p − 1, we have, y − [(k + 1)y − x]p = 1, the minimum solution of which is y = 1, x = (k + 1)y = k + 1. Then, the minimum m in (12) is given by m = qy = q. Note that, from (2.1) with ` = p − 1, we have [(k + 1)y − x]p − x = 1 with the least possible solution y = p − 1 (and (k + 1)y − x = 1). Corollary 2.3. Let p and q be two primes with q > p ≥ 5. Let q = kp+2 for some integer k ≥ 1. Then, q(p − 1) Z(pq) = . 2 Proof. From (2.2) with ` = 2, we have (x − ky)p − 2y = 1, with the minimum solution p−1 q(p−1) y = 2 (and x − ky = 1). This gives m = qy = 2 as one possible solution of (12). p+1 Now, (2.1) with ` = 2 gives 2y − (x − ky)p = 1, with the minimum solution y = 2 (and q(p+1) x = ky + 1). This gives m = qy − 1 = 2 − 1 as another possible solution of (12). Now, q(p+1) q(p−1) since 2 − 1 > 2 , it follows that q(p − 1) Z(pq) = , 2 which we intended to prove. Corollary 2.4. Let p and q be two primes with q > p ≥ 5. Let q = (k + 1)p − 2 for some integer k ≥ 1. Then, q(p − 1) Z(pq) = − 1. 2 Proof. By (2.1) with ` = p − 2, we get [(k + 1)y − x]p − 2y = 1, whose minimum solution p−1 q(p−1) is y = 2 (and x = (k + 1)y − 1). This gives m = qy − 1 = 2 − 1 as one possible solution of (12). Note that, (2.2) with ` = p − 2 gives 2y − [(k + 1)y − x]p = 1, with the minimum solution p+1 q(p+1) y = 2 (and x = (k + 1)y − 1). Corresponding to this case, we get m = qy = 2 as another q(p+1) q(p−1) q(p−1) possible solution of (12). But since 2 > 2 − 1, it follows that Z(pq) = 2 − 1, establishing the theorem. Corollary 2.5. Let p and q be two primes with q > p ≥ 7. Let q = kp+3 for some integer k ≥ 1. Then,   q(p−1) , if 3|(p − 1); Z(pq) = 3  q(p+1) 3 − 1, if 3|(p + 1). Vol. 2 A note on the Pseudo-Smarandache function 11

Proof. From (2.1) and (2.2) with ` = 3, we have respectively

3y − (x − ky)p = 1, (13)

(x − ky)p − 3y = 1. (14)

We now consider the following two possible cases : Case 1 : When 3 divides p − 1. p−1 In this case, the minimum solution is obtained from (14), which is y = 3 (and x−ky = 1). q(p−1) Also, p − 1 is divisible by 2 as well. Therefore, the minimum m in (12) is m = qy = 3 . Case 2 : When 3 divides p + 1. p+1 In this case, (13) gives the minimum solution, which is y = 3 (and x − ky = 1). Note q(p+1) that, 2 divides p + 1. Therefore, the minimum m in (12) m = qy − 1 = 3 − 1. Thus, the theorem is established. Corollary 2.6. Let p and q be two primes with q > p ≥ 7. Let q = (k + 1)p − 3 for some integer k ≥ 1. Then,   q(p+1) , if 3|(p + 1); Z(pq) = 3  q(p−1) 3 − 1, if 3|(p − 1). Proof. From (2.1) and (2.2) with ` = p − 3, we have respectively

[(k + 1)y − x]p − 3y = 1, (15)

3y − [(k + 1)y − x]p = 1. (16)

We now consider the following two cases : Case 1 : When 3 divides p + 1. p+1 In this case, the minimum solution, obtained from(14), is y = 3 (and x = (k + 1)y − q(p+1) 1).Moreover, 2 divides p + 1. Therefore, the minimum m in (12) is m = qy = 3 . Case 2 : When 3 divides p − 1. p−1 In this case, the minimum solution, obtained from (13), is y = 3 (and x = (k + 1)y − 1). q(p−1) Moreover, 2 divides p − 1. Therefore, the minimum m in (12) is m = qy = 3 − 1. Corollary 2.7. Let p and q be two primes with q > p ≥ 7. Let q = kp+4 for some integer k ≥ 1. Then,   q(p−1) , if 4|(p − 1); Z(pq) = 4  q(p+1) 4 − 1, if 4|(p + 1). Proof. From (2.1) and (2.2) with ` = 4, we have respectively

4y − (x − ky)p = 1, (17)

(x − ky)p − 4y = 1. (18)

Now, for any prime p ≥ 7, exactly one of the following two cases can occur : Either p − 1 is divisible by 4, or p + 1 is divisible by 4. We thus consider the two possibilities separately below: Case 1 : When 4 divides p − 1. 12 A.A.K. Majumdar No. 3

p−1 In this case, the minimum solution is obtained from (18), is y = 4 (and x = ky + 1). q(p−1) Therefore, the minimum m in (12) is is m = qy = 4 . Case 2 : When 4 divides p + 1. p+1 In this case, (17) gives the minimum solution, which is y = 4 (and x = ky+1). Therefore, q(p+1) the minimum m in (12) m = qy − 1 = 4 − 1. Corollary 2.8. Let p and q be two primes with q > p ≥ 7. Let q = (k + 1)p − 4 for some integer k ≥ 1. Then,   q(p+1) , if 4|(p + 1); Z(pq) = 4  q(p−1) 4 − 1, if 4|(p − 1). Proof. From (2.1) and (2.2) with ` = p − 4, we have respectively

[(k + 1)y − x]p − 4y = 1, (19)

4y − [(k + 1)y − x]p = 1. (20)

We now consider the following two cases which are the only possibilities (as noted in the proof of Corollary 2.7). Case 1 : When 4 divides p + 1. p+1 In this case, the minimum solution obtained from (20) is y = 4 (and x = (k + 1)y − q(p+1) 1).Therefore, the minimum m in (12) is m = qy = 4 . Case 2 : When 4 divides p − 1. p−1 In this case, the minimum solution, obtained from (19),is y = 4 (and x = (k + 1)y − q(p−1) 1).Therefore, the minimum m in (12) is m = qy = 4 − 1. Corollary 2.9 . Let p and q be two primes with q > p ≥ 11. Let q = kp + 5 for some integer k ≥ 1. Then,   q(p−1) , if 5|(p − 1);  5   q(2a + 1) − 1, if p = 5a + 2; Z(pq) =  q(2a + 1), if p = 5a + 3;   q(p+1) 5 − 1, if 5|(p + 1). Proof. From (2.1) and (2.2) with ` = 5, we have respectively

5y − (x − ky)p = 1, (21)

(x − ky)p − 5y = 1. (22)

Now, for any prime p ≥ 7, exactly one of the following four cases occur: Case 1 : When p is of the form p = 5a + 1 for some integer a ≥ 2. In this case, 5 divides p − 1. Then, the minimum solution is obtained from (22) which is p−1 q(p−1) y = 5 (and x − ky = 1). Therefore, the minimum m in (12) is m = qy = 5 . Case 2 : When p is of the form p = 5a + 2 for some integer a ≥ 2. In this case, from (21) and (22), we get respectively

1 = 5y − (x − ky)(5a + 2) = 5[y − (x − ky)a] − 2(x − ky), (23) Vol. 2 A note on the Pseudo-Smarandache function 13

1 = (x − ky)(5a + 2) − 5y = 2(x − ky) − 5[y − (x − ky)a] (24)

Clearly, the minimum solution is obtained from (23), which is y − (x − ky)a = 1, x − ky = 2 =⇒ y = 2a + 1 (and x = k(2a + 1) + 2). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(2a + 1) − 1. Case 3 : When p is of the form p = 5a + 3 for some integer a ≥ 2. From (21) and (22), we get 1 = 5y − (x − ky)(5a + 3) = 5[y − (x − ky)a] − 3(x − ky), (25)

1 = (x − ky)(5a + 3) − 5y = 3(x − ky) − 5[y − (x − ky)a]. (26)

The minimum solution is obtained from (27) as follows : y − (x − ky)a = 1, x − ky = 2 =⇒ y = 2a + 1 (and x = k(2a + 1) + 2). Hence, in this case, the minimum m in (12) is m = qy = q(2a + 1). Case 4 : When p is of the form p = 5a + 4 for some integer a ≥ 2. In this case, 5 divides p + 1. Then, the minimum solution is obtained from (21), which is p+1 q(p+1) y = 5 (and x − ky = 1). Therefore, the minimum m in (12) is m = qy − 1 = 5 − 1. Corollary 2.10. Let p and q be two primes with q > p ≥ 11. Let q = (k + 1)p − 5 for some integer k ≥ 1. Then,   q(p−1) − 1, if 5|(p − 1);  5   q(2a + 1), if p = 5a + 2; Z(pq) =  q(2a + 1) − 1, if p = 5a + 3;   q(p+1) 5 , if 5|(p + 1).

Proof. From (2.1) and (2.2) with ` = p − 5, we have respectively

[(k + 1)y − x]p − 5y = 1, (27)

5y − [(k + 1)y − x]p = 1. (28)

As in the proof of Corollary 2.9, we consider the following four possibilities : Case 1 : When p is of the form p = 5a + 1 for some integer a ≥ 2. In this case, 5 divides p − 1. Then, the minimum solution is obtained from (27), which is p−1 q(p−1) y = 5 (and x = (k+1)y−1). Therefore, the minimum m in (12) is m = qy−1 = 5 −1. Case 2 : When p is of the form p = 5a + 2 for some integer a ≥ 2. In this case, from (27) and (28), we get respectively

1 = [(k + 1)y − x](5a + 2) − 5y = 2[(k + 1)y − x] − 5[y − a(k + 1)y − x], (29)

1 = 5y − [(k + 1)y − x](5a + 2) = 5[y − a(k + 1)y − x] − 2[(k + 1)y − x]. (30)

Clearly, the minimum solution is obtained from (30), which is y − a(k + 1)y − x = 1, (k + 1)y − x = 2 =⇒ y = 2a + 1 (and x = (k + 1)(2a + 1) − 2). Hence, in this case, the minimum m in (12) is m = qy = q(2a + 1). Case 3 : When p is of the form p = 5a + 3 for some integer a ≥ 2. 14 A.A.K. Majumdar No. 3

In this case, from (27) and (28), we get respectively

1 = [(k + 1)y − x](5a + 3) − 5y = 3[(k + 1)y − x] − 5[y − a(k + 1)y − x], (31)

1 = 5y − [(k + 1)y − x](5a + 3) = 5[y − a(k + 1)y − x] − 3[(k + 1)y − x]. (32)

The minimum solution is obtained from (31) as follows : y − a(k + 1)y − x = 1, (k + 1)y − x = 2 =⇒ y = 2a + 1 (and x = (k + 1)(2a + 1) − 2). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(2a + 1) − 1. Case 4 : When p is of the form p = 5a + 4 for some integer a ≥ 2. In this case, 5 divides p + 1. Then, the minimum solution is obtained from (28), which is p+1 q(p+1) y = 5 (and x = (k + 1)y − 1). Therefore, the minimum m in (12) is m = qy = 5 . Corollary 2.11. Let p and q be two primes with q > p ≥ 13. Let q = kp + 6 for some integer k ≥ 1. Then,   q(p−1) , if 6|(p − 1); Z(pq) = 6  q(p+1) 6 − 1, if 6|(p + 1). Proof. From (2.1) and (2.2) with ` = 6, we have respectively

6y − (x − ky)p = 1, (33)

(x − ky)p − 6y = 1. (34)

Now, for any prime p ≥ 13, exactly one of the following two cases can occur : Either p − 1 is divisible by 6, or p + 1 is divisible by 6. We thus consider the two possibilities separately below : Case 1 : When 6 divides p − 1. p−1 In this case, the minimum solution, obtained from (34), is y = 6 (and x = ky + 1). q(p−1) Therefore, the minimum m in (12) is m = qy = 6 . Case 2 : When 6 divides p + 1. p+1 In this case, (33) gives the minimum solution, which is y = 6 (and x = ky+1). Therefore, q(p+1) the minimum m in (12) is m = qy − 1 = 6 − 1. Corollary 2.12. Let p and q be two primes with q > p ≥ 13. Let q = (k + 1)p − 6 for some integer k ≥ 1. Then,   q(p+1) , if 6|(p + 1); Z(pq) = 6  q(p−1) 6 − 1, if 6|(p − 1).

Proof. From (2.1) and (2.2) with ` = p − 6, we have respectively

[(k + 1)y − x]p − 6y = 1, (35)

6y − [(k + 1)y − x]p = 1. (36)

We now consider the following two cases which are the only possibilities (as noted in the proof of Corollary 2.11) : Case 1 : When 6 divides p + 1. Vol. 2 A note on the Pseudo-Smarandache function 15

p+1 In this case, the minimum solution, obtained from (36), is y = 6 (and x = (k + 1)y − 1). q(p+1) Therefore, the minimum m in (12) is m = qy = 6 . Case 2 : When 6 divides p − 1. p−1 Here, the minimum solution is obtained from (35), which is y = 6 (and x = (k +1)y −1). q(p−1) Therefore, the minimum m in (12) is m = qy = 6 − 1. Corollary 2.13. Let p and q be two primes with q > p ≥ 13. Let q = kp + 7 for some integer k ≥ 1. Then,   q(p−1)  7 , if 7|(p − 1);    q(3a + 1) − 1, if p = 7a + 2;   q(2a + 1) − 1, if p = 7a + 3; Z(pq) =  q(2a + 1), if p = 7a + 4;    q(3a + 2), if p = 7a + 5;   q(p+1) 7 − 1, if 7|(p + 1). Proof. From (2.1) and (2.2) with ` = 7, we have respectively

7y − (x − ky)p = 1, (37)

(x − ky)p − 7y = 1. (38)

Now, for any prime p ≥ 11, exactly one of the following six cases occur : Case 1 : When p is of the form p = 7a + 1 for some integer a ≥ 2. In this case, 7 divides p − 1. Then, the minimum solution is obtained from (38), which is p−1 y = 7 (and x − ky = 1). q(p−1) Therefore, the minimum m in (12) is m = qy = 7 . Case 2 : When p is of the form p = 7a + 2 for some integer a ≥ 2. In this case, from (37) and (38), we get respectively

1 = 7y − (x − ky)(7a + 2) = 7[y − (x − ky)a] − 2(x − ky), (39)

1 = (x − ky)(7a + 2) − 7y = 2(x − ky) − 7[y − (x − ky)a]. (40)

Clearly, the minimum solution is obtained from (39), which is y − (x − ky)a = 1, x − ky = 3 =⇒ y = 3a + 1 (and x = k(3a + 1) + 3). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(3a + 1) − 1. Case 3 : When p is of the form p = 7a + 3 for some integer a ≥ 2. Here, from (37) and (38), 1 = 7y − (x − ky)(7a + 3) = 7[y − (x − ky)a] − 3(x − ky), (41)

1 = (x − ky)(7a + 3) − 7y = 3(x − ky) − 7[y − (x − ky)a]. (42)

The minimum solution is obtained from (41) as follows: y − (x − ky)a = 1, x − ky = 2 =⇒ y = 2a + 1(and x = k(2a + 1) + 2). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(2a + 1) − 1. 16 A.A.K. Majumdar No. 3

Case 4 : When p is of the form p = 7a + 4 for some integer a ≥ 2. In this case, from (37) and (38), we get respectively

1 = 7y − (x − ky)(7a + 4) = 7[y − (x − ky)a] − 4(x − ky), (43)

1 = (x − ky)(7a + 4) − 7y = 4(x − ky) − 7[y − (x − ky)a]. (44)

Clearly, the minimum solution is obtained from (44), which is y − (x − ky)a = 1, x − ky = 2 =⇒ y = 2a + 1(and x = k(2a + 1) + 2). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(2a + 1) + 1. Case 5 : When p is of the form p = 7a + 5 for some integer a ≥ 2. From (37) and (38), we have 1 = 7y − (x − ky)(7a + 5) = 7[y − (x − ky)a] − 5(x − ky), (45)

1 = (x − ky)(7a + 5) − 7y = 5(x − ky) − 7[y − (x − ky)a]. (46)

The minimum solution is obtained from (46), which is y − (x − ky)a = 2, x − ky = 3 =⇒ y = 3a + 2(and x = k(3a + 2) + 3). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(3a + 2). Case 6 : When p is of the form p = 7a + 6 for some integer a ≥ 2. In this case, 7 divides p+1 p + 1. Then, the minimum solution is obtained from (37), which is y = 7 (and x − ky = 1). q(p+1) Therefore, the minimum m in (12) is m = qy − 1 = 7 − 1. Corollary 2.14. Let p and q be two primes with q > p ≥ 13. Let q = (k + 1)p − 7 for some integer k ≥ 1. Then,   q(p−1)  7 , if 7|(p − 1);    q(3a + 1), if p = 7a + 2;   q(2a + 1), if p = 7a + 3; Z(pq) =  q(2a + 1) − 1, if p = 7a + 4;    q(3a + 2) − 1, if p = 7a + 5;   q(p+1) 7 , if 7|(p + 1). Proof. From (2.1) and (2.2) with ` = 7, we have respectively

[(k + 1)y − x]p − 7y = 1, (47)

7y − [(k + 1)y − x]p = 1. (48)

We now consider the following six possibilities: Case 1 : When p is of the form p = 7a + 1 for some integer a ≥ 2. In this case, 7 divides p−1 p−1. Then, the minimum solution is obtained from (47), which is y = 7 (andx = (k+1)y−1). q(p−1) Therefore, the minimum m in (12) is m = qy − 1 = 7 − 1. Case 2 : When p is of the form p = 7a + 2 for some integer a ≥ 2. In this case, from (47) and (48), we get respectively

1 = [(k + 1)y − x](7a + 2) − 7y = 2[(k + 1)y − x] − 7[y − a{(k + 1)y − x}], (49) Vol. 2 A note on the Pseudo-Smarandache function 17

1 = 7y − [(k + 1)y − x](7a + 2) = 7[y − a{(k + 1)y − x}] − 2[(k + 1)y − x]. (50)

Clearly, the minimum solution is obtained from (50), which is y − a{(k + 1)y − x} = 1, (k + 1)y − x = 3 =⇒ y = 3a + 1(and x = (k + 1)(3a + 1) − 3). Hence, in this case, the minimum m in (12) is m = qy = q(3a + 1). Case 3 : When p is of the form p = 7a + 3 for some integer a ≥ 2. Here, from (47) and (48),

1 = [(k + 1)y − x](7a + 3) − 7y = 3[(k + 1)y − x] − 7[y − a{(k + 1)y − x}], (51)

1 = 7y − [(k + 1)y − x](7a + 3) = 7[y − a{(k + 1)y − x}] − 3[(k + 1)y − x]. (52)

Then, (52) gives the minimum solution, which is: y − a{(k + 1)y − x} = 1, (k + 1)y − x = 2 =⇒ y = 2a + 1(and x = (k + 1)(2a + 1) − 2). Hence, in this case, the minimum m in (12) is m = qy = q(2a + 1). Case 4 : When p is of the form p = 7a + 4 for some integer a ≥ 2. Here, from (47) and (48),

1 = [(k + 1)y − x](7a + 4) − 7y = 4[(k + 1)y − x] − 7[y − a{(k + 1)y − x}], (53)

1 = 7y − [(k + 1)y − x](7a + 4) = 7[y − a{(k + 1)y − x}] − 4[(k + 1)y − x]. (54)

Clearly, the minimum solution is obtained from (53) as follows: y − a{(k + 1)y − x} = 1, (k + 1)y − x = 2 =⇒ y = 2a + 1(and x = (k + 1)(2a + 1) − 2). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(2a + 1) − 1. Case 5 : When p is of the form p = 7a + 5 for some integer a ≥ 2. In this case, from (47) and (48), we get respectively

1 = [(k + 1)y − x](7a + 5) − 7y = 5[(k + 1)y − x] − 7[y − a{(k + 1)y − x}], (55)

1 = 7y − [(k + 1)y − x](7a + 5) = 7[y − a{(k + 1)y − x}] − 5[(k + 1)y − x]. (56)

Then, (55) gives the following minimum solution: y − a{(k + 1)y − x} = 2, (k + 1)y − x = 3 =⇒ y = 3a + 2 (and x = (k + 1)(3a + 2) − 3). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(3a + 2) − 1. Case 6 : When p is of the form p = 7a+6 for some integer a ≥ 2. In this case, 7 divides p+1. p+1 Then, the minimum solution is obtained from (48), which is y = 7 (and x = (k + 1)y − 1). q(p+1) Therefore, the minimum m in (12) is m = qy = 7 . Corollary 2.15. Let p and q be two primes with q > p ≥ 13. Let q = kp + 8 for some integer k ≥ 1. Then,   q(p−1) , if 8|(p − 1);  8   q(3a + 1), if p = 8a + 3; Z(pq) =  q(3a + 2) − 1, if p = 8a + 5;   q(p+1) 8 − 1, if 8|(p + 1). 18 A.A.K. Majumdar No. 3

Proof. From (2.1) and (2.2) with ` = 8, we have respectively

8y − (x − ky)p = 1, (57)

(x − ky)p − 8y = 1. (58)

Now, for any prim p ≥ 13, exactly one of the following four cases occur: Case 1 : When p is of the form p = 8a + 1 for some integer a ≥ 2. In this case, 8 divides p − 1. Then, the minimum solution is obtained from (58), which is p−1 y = 8 (andx − ky = 1). q(p−1) Therefore, the minimum m in (12) is m = qy = 8 . Case 2 : When p is of the form p = 8a + 3 for some integer a ≥ 2. In this case, from (57) and (58), we get respectively

1 = 8y − (x − ky)(8a + 3) = 8[y − (x − ky)a] − 3(x − ky), (59)

1 = (x − ky)(8a + 3) − 8y = 3(x − ky) − 8[y − (x − ky)a]. (60)

Clearly, the minimum solution is obtained from (60), which is y − (x − ky)a = 1, x − ky = 3 =⇒ y = 3a + 1 (and x = k(3a + 1) + 3). Hence, in this case, the minimum m in (12) is m = qy = q(3a + 1). Case 3 : When p is of the form p = 8a + 5 for some integer a ≥ 2. From (57) and (58), We get

1 = 8y − (x − ky)(8a + 5) = 8[y − (x − ky)a] − 5(x − ky), (61)

1 = (x − ky)(8a + 5) − 8y = 5(x − ky) − 8[y − (x − ky)a]. (62)

The minimum solution is obtained from (61) as follows: y − (x − ky)a = 2, x − ky = 3 =⇒ y = 3a + 2 (and x = k(3a + 2) + 3). Hence, in this case, the minimum m in (12) is m = qy − 1 = q(3a + 2) − 1. Case 4 : When p is of the form p = 8a + 7 for some integer a ≥ 2. In this case, 8 divides p − 1. Then, the minimum solution is obtained from (57), which is p+1 q(p+1) y = 8 (andx − ky = 1). Therefore, the minimum m in (12) is m = qy − 1 = 8 − 1. Corollary 2.16. Let p and q be two primes with q > p ≥ 13. Let q = (k + 1)p − 8 for some integer k ≥ 1. Then,   q(p−1) , if 8|(p − 1);  8   q(3a + 1) − 1, if p = 8a + 3; Z(pq) =  q(3a + 2), if p = 8a + 5;   q(p+1) 8 , if 8|(p + 1). Proof. From (2.1) and (2.2) with ` = p − 8, we have respectively

[(k + 1)y − x]p − 8y = 1, (63)

8y − [(k + 1)y − x]p = 1. (64) Vol. 2 A note on the Pseudo-Smarandache function 19

We now consider the four possibilities that may arise: Case 1 : When p is of the form p = 8a + 1 for some integer a ≥ 2. In this case, 8 divides p − 1. Then, the minimum solution is obtained from (63), which is p−1 q(p−1) y = 8 (and x = (k + 1)y − 1). Therefore, the minimum m in (12) is m = qy − 1 = 8 − 1. Case 2 : When p is of the form p = 8a + 3 for some integer a ≥ 2. In this case, from (63) and (64), we get respectively

1 = [(k + 1)y − x](8a + 3) − 8y = 2[(k + 1)y − x] − 8[y − a{(k + 1)y − x}], (65)

1 = 8y − [(k + 1)y − x](8a + 3) = 8[y − a{(k + 1)y − x}] − 3[(k + 1)y − x]. (66)

Clearly, the minimum solution is obtained from (65), which is

y − a{(k + 1)y − x} = 1, (k + 1)y − x = 3 =⇒ y = 3a + 1(andx = (k + 1)(3a + 1) − 3).

Hence, in this case, the minimum m in (12) is m = qy = q(3a + 1) − 1. Case 3 : When p is of the form p = 8a + 5 for some integer a ≥ 1. In this case, from (63) and (64), we get respectively

1 = [(k + 1)y − x](8a + 5) − 8y = 5[(k + 1)y − x] − 8[y − a{(k + 1)y − x}], (67)

1 = 8y − [(k + 1)y − x](8a + 5) = 8[y − a{(k + 1)y − x}] − 5[(k + 1)y − x]. (68)

The minimum solution is obtained from (68) as follows:

y − a{(k + 1)y − x} = 2, (k + 1)y − x = 3 =⇒ y = 3a + 2(andx = (k + 1)(3a + 2) − 3).

Hence, in this case, the minimum m in (12) is m = qy = q(3a + 2). Case 4 : When p is of the form p = 8a + 7 for some integer a ≥ 2. In this case, 8 divides p + 1. Then, the minimum solution is obtained from (64), which is p+1 q(p+1) y = 8 (and x = (k + 1)y − 1). Therefore, the minimum m in (12) is m = qy = 8 . We now consider the case when n is a composite number. Let m0(m0+1) Z(n) = m0 for some integer m0 ≥ 1. Then, n divides 2 . We now consider the following two cases that may arise :

Case 1 : m0 is even (so that m0 + 1 is odd). m0 (1) Let n be even. In this case, n does not divide 2 , for otherwise, n|m n|m (m + 1) 0 =⇒ 0 0 =⇒ Z(n) ≤ (m − 1). 2 2 0

(2) Let n be odd. In such a case, n does not divide m0.

Case 2 : m0 is odd (so that m0 + 1 is even).

(1) Let n be even. Then, n does not divide m0.

(2) Let n be odd. Here, n does not divide m0, for n|m (m − 1) n|m =⇒ 0 0 =⇒ Z(n) ≤ (m − 1). 0 2 0 20 A.A.K. Majumdar No. 3

Thus, if n is a composite number, n does not divide m0. Now let

α1 α2 αi αi+1 αs n = p1 p2 ··· pi pi+1 ··· ps

be the representation of n in terms of its distinct prime factors p1, p2, ··· pi, pi+1, ··· ps, not

necessarily ordered. Then, one of m0 and m0 + 1 is of the form

β β1 β2 βi βi+1 βs 2 p1 p2 ··· pi qi+1 ··· qs

for some 1 ≤ i < s; βj ≥ αj for 1 ≤ j < i, and the other one is of the form

γi+1 γs γs+1 γu pi+1 ··· ps , rs+1 ··· ru γj ≥ αj

for i + 1 ≤ j < s; where qi+1, ··· qs and rs+1, ··· ru are all distinct primes, not necessarily ordered.

§3. Some Observations

Some observations about the Pseudo-Smarandache Function are given below : Remark 3.1. Kashihara raised the following questions (see Problem 7 in [1]) : (1) Is there any integer n such that Z(n) > Z(n + 1) > Z(n + 2) > Z(n + 3)? (2) Is there any integer n such that Z(n) < Z(n + 1) < Z(n + 2) < Z(n + 3)? The following examples answer the questions in the aﬃrmative:

(1) Z(256) = 511 > 256 = Z(257) > Z(258) = 128 > 111 = Z(259) > Z(260) = 39,

(2) Z(159) = 53 < 64 = Z(160) < Z(161) = 69 < 80 = Z(162) < Z(163) = 162.

These examples show that even ﬁve consecutive increasing or decreasing terms are available in the sequence {Z(n)}. Remark 3.2 Kashihara raises the following question (see Problem 5 in [1]) : Given any integer

m0 ≥ 1, how many n are there such that Z(n) = m0?

Given any integer m0\3, let

−1 Z (m0) = {n : n ∈ N,Z(n) = m0}, (2.3)

with Z−1(1) = {1},Z−1(2) = {3}. (2.4)

Thus, for example, Z−1(8) = {8, 12, 18, 36}. By Lemma 2.1, m (m + 1) n ≡ 0 0 ∈ Z−1(m ). max 2 0 −1 This shows that the set Z (m0) is non-empty; moreover, nmax is the biggest element of −1 −1 −1 Z (m0), so that Z (m0) is also bounded. Clearly, n ∈ Z (m0) only if n divides f(m0) ≡ Vol. 2 A note on the Pseudo-Smarandache function 21

m0(m0 + 1)/2. This is a necessary condition, but is not suﬃcient. For example, 4|36 ≡ f(8) but 4 ∈/ Z−1(8). The reason is that Z(n) is not bijective. Let X∞ Z−1 ≡ Z−1(m) m=1 −1 −1 Let n ∈ Z . Then, there is one and only one mo such that n ∈ Z (m0), that is, there is one

and only one mo such that Z(n) = m0. −1 However, we have the following result whose proof is almost trivial : n ∈ Z (m0)(n 6= 1, 3) if and only if the following two conditions are satisﬁed

(1) n divides m0(m0 + 1)/2,

(2) n does not divide m(m + 1)/2 for any m with 3 ≤ m ≤ m0 − 1. Since 4|28 ≡ f(7), it therefore follows that 4 ∈/ Z−1(8).

Given any integer m0 ≥ 1, let C(m0) be the number of integers n such that Z(n) = m0, that −1 is, C(m0) denotes the number of elements of Z (m0). Then,

1 ≤ C(m0) ≤ d(m0(m0 + 1)/2) − 1 form0 ≥ 3; C(1) = 1,C(2) = 2,

where, for any integer n, d(n) denotes the number of divisors of n including 1 and n. Now, let p ≥ 3 be a prime. Since, by Lemma 1.2, Z(p) = p − 1, we see that p ∈ Z−1(p − 1) for all p ≥ 3. Let n ∈ Z−1(p − 1). Then, n divides p(p − 1)/2. This shows that n must divide p, for otherwise p − 1 (p − 1)(p − 2) n| ⇒ n| ⇒ Z(n) ≤ p − 2, 2 2 contradicting the assumption. Thus, any element of Z−1(p−1) is a multiple of p. In particular, p is the minimum element of Z−1(p − 1). Thus, if p ≥ 5 is a prime, then Z−1(p − 1) contains

at least two elements, namely, p and p(p − 1)/2. Next, let p be a prime factor of m0(m0 + 1)/2. −1 Since, by Lemma 1.2, Z(p) = p − 1, we see that p ∈ Z (m0) if and only if p − 1 ≥ m0, that

is, if and only if p ≥ m0 + 1. Remark 3.3. Ibstedt[2] provides a table of values of Z(n) for 1 ≤ n ≤ 1000. A closer look at these values reveal some facts about the values of Z(n). These observations are given in the conjectures below, followed by discussions in each case. Conjecture 1. Z(n) = 2n − 1 if and only if n = 2k for some integer k ≥ 0. Let, for some integer n ≥ 1,

Z(n) = m0, where m0 = 2n − 1.

Note that the conjecture is true for n = 1 (with k = 0). Also, note that n must be composite. m0(m0+1) Now, since m0 = 2n − 1, and since n| 2 , it follows that m0+1 n does not divide m0, and n| 2 ; moreover, by virtue of the deﬁnition of Z(n), n does not m+1 divide m0, and n| 2 for all 1 ≤ m ≤ m0 − 1. Let

Z(2n) = m1.

m0+1 2(m0+1) (2m0+1)+1 We want to show that m1 = 2mo + 1. Since n| 2 , it follows that 2n| 2 = 2 ; moreover, 2n does not divide 2(m + 1) (2m + 1) + 1 = 2 2 22 A.A.K. Majumdar No. 3

for all 1 ≤ m ≤ m0 − 1. Thus, 2 m1 = 2m0 + 1 = 2(2n − 1) + 1 = 2 n − 1. All these show that Z(n) = 2n − 1 ⇒ Z(2n) = 22n − 1.

Continuing this argument, we see that

Z(n) = 2n − 1 ⇒ Z(2kn) = 2k+1n − 1.

Since Z(1) = 1, it then follows that Z(2k) = 2k+1 − 1. Conjecture 2. Z(n) = n − 1 if and only if n = pk for some prime p ≥ 3 and integer k ≥ 1. Let, for some integer n ≥ 2,

Z(n) = m0, where m0 = n − 1.

Then, 2|m0 and n|(m0 + 1); moreover, n does not divide m + 1 for any 1 ≤ m ≤ m0 − 1. Let 2 Z(n ) = m1.

Since n|(m0 + 1), it follows that

2 2 2 n |(m0 + 1) = (m0 + 2m0) + 1;

2 2 2 moreover, n does not divide|(m + 1) = (m + 2m) + 1 for all 1 ≤ m ≤ m0 − 1. Thus, 2 2 2 m1 = m0 + 2m0 = (n − 1) + 2(n − 1) = n − 1,

so that(since 2|m0 ⇒ 2|m1)

Z(n) = n − 1 ⇒ Z(n2) = n2 − 1.

Continuing this argument, we see that

Z(n) = n − 1 ⇒ Z(n2k) = n2k − 1.

Next, let 2k+1 Z(n ) = m2 for some integer k ≥ 1.

Since n|(m0 + 1), it follows that

2k+1 2k+1 2k+1 n |(m0 + 1) = [(m0 + 1) − 1] + 1;

moreover, n2k+1 does not divide |(m + 1)2k+1 = [(m + 1)2k+1 − 1] + 1

for all 1 ≤ m ≤ m0 − 1. Thus,

2k+1 2k+1 m2 = (m0 + 1) − 1 = n − 1, Vol. 2 A note on the Pseudo-Smarandache function 23

so that (since 2|m0 ⇒ 2|m2)

Z(n) = n − 1 ⇒ Z(n2k+1) = n2k+1 − 1.

All these show that Z(n) = n − 1 ⇒ Z(nk) = nk − 1.

Finally, since Z(p) = p − 1 for any prime p ≥ 3, it follows that Z(pk) = pk − 1. Conjecture 3. If n is not of the form 2k for some integer k ≥ 0, then Z(n) < n. First note that, we can exclude the possibility that Z(n) = n, because

n(n + 1) n(n − 1) n| ⇒ n| ⇒ Z(n) ≤ n − 1. 2 2 So, let

Z(n) = m0 with m0 > n.

Note that, n must be a composite number, not of the form pk (p ≥ 3 is prime, k ≥ 0). Let

m0 = an + b for some integers a ≥ 1, 1 ≤ b ≤ n1.

Then, 2 m0(m0 + 1) = (an + b)(an + b + 1) = n(a n + 2ab + a) + b(b + 1).

Therefore,

n|m0(m0 + 1) if and only if b + 1 = n.

But, by Conjecture 1, b + 1 = n leads to the case when n is of the form 2k. Remark 3.4. Kashihara proposes (see Problem 4(a) in [1]) to ﬁnd all the values of n such that Z(n) = Z(n + 1). In this connection, we make the following conjecture : Conjecture 4. For any integer n ≥ 1, Z(n) 6= Z(n + 1). Let

Z(n) = Z(n + 1) = m0 for some n ∈ N, m0 ≥ 1. (69)

Then, neither n nor n + 1 is a prime. To prove this, let n = p, where p is a prime. Then, by Lemma 1.2, Z(n) = Z(p) = p − 1.

p(p − 1) n + 1 = p + 1 does not divide ⇒ Z(n + 1) 6= p − 1 = Z(n). 2 Similarly, it can be shown that n + 1 is not a prime. Thus, both n and n + 1 are composite numbers.

From (68), we see that both n and n + 1 divide m0(m0 + 1)/2. Let

m (m + 1) 0 0 = an for some integer a ≥ 1. 2

Since n + 1 divides m0(m0 + 1) and since n + 1 does not divide n, it follows that n + 1 must divide a. So, let a = b(n + 1) for some integer b ≥ 1. 24 A.A.K. Majumdar No. 3

Then, m (m + 1) 0 0 = abn(n + 1), 2 which shows that m (m + 1) n(n + 1) must divide 0 0 . (70) 2 From (69), we see that

Z(n(n + 1)) ≤ m0,

which, together with Lemma 1.5 (that Z(n(n + 1)) ≥ Z(n)), gives

Z(n(n + 1)) = m0. (71)

From (70), we see that

m (m + 1) n(n + 1) m (m + 1) n(n + 1) n(n + 1) 0 0 ⇒ | 0 0 ⇒ Z( ) ≤ m . 2 2 2 2 0

n(n+1) Thus, by virtue of Lemma 2.1, Z( 2 ) = n ≤ m0 = Z(n). It can easily be veriﬁed that neither n nor n + 1 can be of the form 2k. Thus, if Conjecture 3 is true then Conjecture 4 is also true. Remark 3.5. An integer n > 0 is called f-perfect if

Xk n = f(di), i=1

where d1 ≡ 1, d2, . . . , dk are the proper divisors of n, and f is an arithmetical function. In particular, n is Pseudo-Smarandache perfect if

Xk n = Z(di). i=1

In [4], Ashbacher reports that the only Pseudo-Smarandache perfect numbers less than 1, 000, 000 are n = 4, 6, 471544. However, since n = 471544 is of the form n = 8p with p = 58943, its only perfect divisors are 1, 2, 4, 8, p, 2p and 4p. Since 8|(p + 1) = 58944, it follows from Lemma 1.2, Theorem 2.1 and Theorem 2.7 that

Z(p) = p − 1,Z(2p) = p, Z(4p) = p,

so that Xk n = 471544 > Z(di), i=1 so that n = 471544 is not Pseudo-Smarandache perfect. Vol. 2 A note on the Pseudo-Smarandache function 25 References

[1] Kashihara, Kenichiro, Comments and Topics on Smarandache Notions and Problems, USA, Erhus University Press, 1996. [2] Ibstedt, Henry, Surﬁng on the Ocean of Numbers-A Few Smarandache Notions and Similar Topics, USA, Erhus University Press, 1997. [3] Gioia, Anthony A, The Theory of Numbers–An Introduction, NY, USA, Dover Pub- lications Inc., 2001. [4] Ashbacher, Charles, On Numbers that are Pseudo-Smarandache and Smarandache perfect, Smarandache Notions Journal, 14(2004), pp. 40–41. Scientia Magna Vol. 2 (2006), No. 3, 26-30

Primes in the Smarandache deconstructive sequence

Shyam Sunder Gupta Chief Bridge Engineer, North Central Railway, Allahabad, India

Abstract In this article, we present the results of investigation of ﬁrst 10000 terms of Smarandache Deconstructive Sequence and report some new primes and other results found from the sequence. Keywords The Smarandache deconstructive sequence, initial digits, trailing digits.

Introduction

The Smarandache Deconstructive sequence of integers [1] is constructed by sequentially repeating the digits 1 to 9 as follows: 1, 23, 456, 7891, 23456, 789123, ······ Kashihara [2] asked: How many primes are there in the sequence. Ashbacher [3] explored this question and raised some more questions, which were studied and answered by Henry Ibstedt [4]. Let us call the sequence mentioned above as Smarandache Deconstructive sequence of the first kind (SDS-I) because a similar Deconstructive sequence can be constructed by sequentially repeating the digits 0 to 9 as follows [4]. 0, 12, 345, 6789, 01234, 567890, ······ The Smarandache Deconstructive sequence of integers [1] is constructed by sequentially repeating the digits 1 to 9 as follows: 1, 23, 456, 7891, 23456, 789123, ······ Kashihara [2] asked: How many primes are there in the sequence. Ashbacher [3] explored this question and raised some more questions, which were studied and answered by Henry Ibstedt [4]. Let us call the sequence mentioned above as Smarandache Deconstructive sequence of the first kind (SDS-I) because a similar Deconstructive sequence can be constructed by sequentially repeating the digits 0 to 9 as follows [4]. 0, 12, 345, 6789, 01234, 567890, ······ This can be termed as the Smarandache deconstructive sequence of the second kind (SDS- II). In this paper, we report the primes found in both the sequence after checking the first 10000 terms of both these sequence. This can be termed as the Smarandache deconstructive sequence of the second kind (SDS- II). Vol. 2 Primes in the Smarandache deconstructive sequence 27

In this paper, we report the primes found in both the sequence after checking the first 10000 terms of both these sequence. Primes in the Smarandache Deconstructive Sequence of first kind: The following 13 primes in the Smarandache Deconstructive sequence of first kind have been reported earlier [5] [6]. 23, 4567891, 23456789, ······ These are 2, 7, 8, 10, 17, 20, 25, 28, 31, 38, 61, 62 and 355-th term of the sequence and are given in Table-1 below:

Table 1

Term Prime 2 23 7 4567891 8 23456789 10 1234567891 17 23456789123456789 20 23456789123456789123 25 4567891234567891234567891 28 1234567891234567891234567891 31 7891234567891234567891234567891 38 23456789123456789123456789123456789123 61 4567891234567891234567891234567891234567891234567891234567891 62 23456789123456789123456789123456789123456789123456789123456789

355 789(123456789)391

Note that (123456789)39 means 123456789 repeated 39 times. On further computation up to 10000 terms of the sequence, we have noted following 3 more primes, namely the term 4690, 4772 and 8162 of the sequence. Since the primality of the term 4690, 4772 and 8162 have not been certiﬁed, so these can be treated as probable primes. These are:

(123456789)5211,

23456789(123456789)529123,

23456789(123456789)906. It may be noted that though there are 12 primes in the first 62 terms of the sequence but only 16 primes in the first 10000 terms of the sequence. So the percentage of primes is reducing significantly which is in accordance with prime number theorem, according to which, 1 the probability that a random chosen number of size n is prime decreases as d (where d is the number of digits of n). Observations on the Smarandache Deconstructive Sequence of first kind: 28 Shyam Sunder Gupta No. 3

From the term of this sequence, it is seen that the trailing digit (units digit) repeats the pattern. 1, 3, 6, 1, 6, 3, 1, 9, 9; ······ Interestingly this sequence is the same as the sequence of digital root of triangular numbers. Similarly initial digit of the element of SDS-I repeats the pattern 1, 2, 4, 7, 2, 7, 4, 2, 1; ······ Table-2 below gives the possible combination of initial and trailing digits of any element of SDS-I.

Table 2

Trailing digits Initial digits 1 1, 4, 7 3 2, 7 6 4, 2 9 2, 1

Since the trailing digits of the term of the SDS-I sequence can be 1, 3, 6 or 9, it is obvious that for an element to be prime, the only possible trailing digits are 1, 3 or 9. If trailing digit is 3, possible initial digits are 2 and 7, but if initial digit is 7 and trailing digit is 3, the number is divisible by 3. Similarly if trailing digit is 9 and initial digit is 1, the number is divisible by 3. The possible combinations of trailing and initial digits for a prime in the sequence are given in Table-3.

Table 3

Trailing digits Initial digits 1 1, 4, 7 3 2 9 2

So there are 5 possibilities out of 9, as the pattern repeat for every 9 elements in the sequence. Out of 13 primes found, 7 ends in 1, 3 ends in 3, 3 end in 9 and primes corresponding to all 5 possibilities are found. The three probable primes found end in 1, 3 and 9 respectively. It is thus clear that the term 3 + 9n, 5 + 9n, 6 + 9n, 9n of the sequence are obviously composite and need not be checked for primality. Only the term 9n + 1, 9n + 2, 9n + 4, 9n + 7 and 9n + 8 need to be checked for primality. Conjecture 1. Every prime except 5 divides some element of the sequence. It is noted that none of the element of the sequence end in 0 or 5. So 5 cannot be a factor of any terms of the sequence. It has been checked that every prime up to 3821 except 5 divides some element of the sequence up to 10000 terms, so it is quite reasonable to conjecture that: every prime except 5 divides some element of the sequence. Can this be proved? Vol. 2 Primes in the Smarandache deconstructive sequence 29

Primes in the Smarandache Deconstructive Sequence of second kind: On computation up to 10000 terms of the sequence, we have noted only 2 primes, namely the term 367 and 567 of the sequence. These are:

(1234567890)361234567,

(1234567890)561234567. Observations on the Smarandache Deconstructive Sequence of second kind: From the terms of this sequence, it is seen that the trailing digit (units digit) repeats the pattern. 0, 2, 5, 9, 4, 0, 7, 5, 4, 4, 5, 7, 0, 4, 9, 5, 2, 0, 9, 9; ······ Similarly initial digit of the element of SDS-II repeats the pattern 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0; ······ Table-4 below gives the possible combination of initial and trailing digits of any element of SDS-II

Table 4

Trailing digits Initial digits 0 0, 5, 8, 3 2 1, 6 4 0, 5, 6, 1 5 3, 5, 8, 0 7 1, 6 9 1, 0, 5, 6

Since the trailing digits of the term of the SDS-II sequence can be 0, 2, 4, 5, 7 or 9, it is obvious that for an element to be prime, the only possible trailing digits are 7 or 9. If trailing digit is 7, possible initial digits are 1 or 6. Similarly if trailing digit is 9, possible initial digits are 0, 1, 5 or 6. If initial digit is 0, 1 or 6 and trailing digit is 9, the number is divisible by 3. So it cannot be prime. The possible combinations of trailing and initial digits for a prime in the sequence are given in Table-5.

Table 5

Trailing digits Initial digits 7 1, 6 9 5

So there are 3 possibilities out of 20, as the pattern repeat for every 20 elements in the sequence. Both the primes in ﬁrst 10000 terms of the sequence end in 7 and initial digit in both primes is 1. It remains to ﬁnd a single prime corresponding to trailing digit 9 and also corresponding to trailing digit 7 with initial digit 6. The only terms needs to be checked for 30 Shyam Sunder Gupta No. 3 primality are 7 + 20n, 12 + 20n and 15 + 20n. It is interesting to note that for every 20 terms of the sequence, only 3 needs to be checked for possible primes, whereas in SDS-I, for every 9 terms of the sequence, 5 terms needs to be checked for possible primes. This gives an indication that if there are n1 possible primes in SDS-I, then in SDS-II, the number of possible primes 3 9 n2 = n1 ∗ ( 20 ) ∗ ( 5 ) = 0.27n1 This explains why the number of primes found in SDS-II is fewer as compared to number of primes found in SDS-I. The time required to search for primes in SDS-I is also correspondingly higher than the time required to search for primes in SDS-II. Conjecture 2. Every prime divides some element of the sequence. It has been checked that every prime up to 2591 divides some element of the sequence up to 10000 terms, so it is again quite reasonable to conjecture that: every prime divides some element of the sequence. Can this be proved?

References

[1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 1993. [2] K. Kashihara, Comments and Topics on Smarandache Notions and Problems, Erhus University Press, Vail, Arizona, 1996. [3] Charles Ashbacher., Some Problems Concerning The Smarandache Deconstructive Se- quence, Journal of Recreational Mathematics, 2(1998), Baywood Publishing Company, Inc. [4] Henry Ibstedt, On a concatenation problem, Smarandache Notions Journal, 13(2002), pp. 96-106. [5] Charles Ashbacher, Pluckings from the tree of Smarandache Sequences and Functions, American Research Press, 1998. [6] Jason Earls, Some Results Concerning The Smarandache Deconstructive Sequence, Smarandache Notions Journal, 14( 2002), pp. 222-226. [7] Sloane, N.J.A., Sequence A007923, A050234 in ” The on line version of the Encyclopedia of Integer Sequences”. http://www.research.att.com/ njas/sequences/. [8] Weisstein, Eric W, ”Smarandache Sequences”, CRC Concise Encyclopedia of Mathe- matics, CRC Press, 1999. Scientia Magna Vol. 2 (2006), No. 3, 31-40

Recursion formulae for Riemann Zeta function and Dirchlet series1

Suling Zhang† and Chaoping Chen‡ †. Department of Basic Courses, Jiaozuo University, Jiaozuo, Henan, China ‡. College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo , Henan, China

Abstract In this paper, some recursion formulae of sums for the Riemann Zeta function and Dirchlet series are obtained through expanding several simple function on [−π, π] or [0, 2π] by using the Dirichlet theorem in Fourier series theory. Keywords Riemann zeta function, Dirichlet series, recurrence formula, Fourier series.

§1. Introduction

It is well-known that the Riemann Zeta function deﬁned by

X∞ 1 ζ(s) = , <(s) > 1 (1) ns n=1 and Dirichlet series X∞ (−1)n−1 D(s) = , <(s) > 1 (2) ns n=1 play very important roles in Analytic Number Theory, and so on. In 1734, Euler gave sum of the following Bernoulli series

X∞ 1 π2 ζ(2) = = . (3) n2 6 n=1

The formula (3) has been studied by many mathematicians and many proofs have been published, for example, see [2]. In 1748, Euler further gave the following general formula

X∞ 1 (−1)k−122k−1π2k ζ(2k) = = B , (4) n2k (2k)! 2k n=1 where B2k(k = 1, 2,...) denotes Bernoulli numbers, deﬁned in [18,19] by

t X∞ tk = B , |t| < 2π. et − 1 k! k k=0

The authors were supported in part by NNSF (#10001016) of CHINA, SF for the Prominent Youth of 1Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, Doctor Fund of Jiaozuo Institute of Technology, CHINA 32 Suling Zhang and Chaoping Chen No. 3

For other proofs concerning formula (4), please refer to the references in this paper, for example, [18] and [21]. In 1999, the paper [9] gave an elementary expression for ζ(2k): Let n ∈ N, then

X∞ 1 = A π2k, (5) n2k k n=1 where 1 1 1 k A = A − A + ··· + (−1)k−2 A + (−1)k−1 k 3! k−1 5! k−2 (2k − 1)! 1 (2k + 1)! k kX−1 (−1)k−i−1 = (−1)k−1 + A . (6) (2k + 1)! (2k − 2i + 1)! i i=1 It is still an open problem to prove irrationality of ζ(2k + 1) for several centuries. Until 1978, R. Ap´ery, a French mathematician, proved that the number ζ(3) is irrational. But one can not generalize his proof to other cases. So, many mathematicians keep much interest in the evaluation of ζ(s) and sums of related series. For some examples, see [10,20,22]. The following formulae involving ζ(2k + 1) were given by Ramanujan, see [22], as follows:

1. If k > 1 and k ∈ N, " # " # 1 X∞ n2k−1 1 X∞ n2k−1 αk ζ(1 − 2k) + = (−β)k ζ(1 − 2k) + , (7) 2 e2nα − 1 2 e2nβ − 1 n=1 n=1

2. if k > 0 and k ∈ N, " # 1 1 X∞ 1 0 = ζ(2k + 1) + (4α)k 2 n2k+1(e2nα − 1) " n=1 # 1 1 X∞ 1 − ζ(2k + 1) + (−4β)k 2 n2k+1(e2nβ − 1) (8) n=1 k+1 [ 2 ] X0 (−1)jπ2jB B £ ¤ + 2j 2k−2j+2 αk−2j+1 + (−β)k−2j+1 , (2j)!(2K − 2J + 2)! j=0

2 P0 where Bj is the j-th Bernoulli number, α > 0 and β > 0 satisfy αβ = π , and means that, when k is an odd number 2m − 1, the last term of the left hand side in (8) is taken as (−1)mπ2mB2 2m . (m!)2 In 1928, Hardy in [6] proved (7). In 1970, E. Grosswald in [3] proved (8). In 1970, E. Grosswald in [4] gave another expression of ζ(2k + 1). In 1983, N.-Y. Zhang in [20] not only proved Ramanujan formulae (7) and (8), but also gave an explicit expression of ζ(2k + 1) as follows:

1. If k is odd, then we have

k+1 [ 2 ] X0 (−1)jπ2jB B ζ(2k + 1) = −2ψ (π) − (2π)2k+1 2j 2k−2j+2 ; (9) −k (2j)!(2k − 2j + 2)! j=0 Vol. 2 Recursion formulae for Riemann Zeta function and Dirchlet series 33

2. if k is even,

k 2π (2π)2k+1 X2 (−1)jπ2jB B ζ(2k + 1) = −2ψ (π) + ψ0 (π) 2j 2k−2j+2 , (10) −k k −k k (2j)!(2k − 2j + 2)! j=0

X∞ 1 where ψ (α) = , and ψ0 (α) is the derivative of ψ (α) with respect to −k n2k+1(e2nα − 1) −k −k n=1 α. There are a lot of literature on calculating of ζ(s), for example, see [2,p.435] and [18,pp.144- 145; p.149; pp.150-151]. As a matter of fact, many other recent investigations and important results on the subject of the Riemannian Zeta function ζ(s) can be found in the papers [11,12,13,14,15,16,17] by H. M. Srivastava, and others. Furthermore, Chapter 4 entitled “Evaluations and Series Representations”of the book [15] contains a rather systematic pre- sentation of much of these recent developments. The aim of this paper is to obtain recursion formulae of sums for the Riemann Zeta function and Dirchlet series through expanding some simple function on [−π, π] or [0, 2π] by using the Dirichlet theorem in Fourier series theory.

§2. Main results and proofs

X∞ 1 X∞ (−1)n−1 Theorem 1. Let δ(s) , and σ(s) , for s > 1. Then we have (2n − 1)s (2n − 1)s n=1 n=1 for k ∈ N (−1)k+1(4k − 1)π2k Xk (−1)k+j+14jπ2k−2j D(2k) = + ζ(2j), (11) 2 · (2k)! (2k − 2j)! j=1 (−1)k+1π2k Xk (−1)k+j+1π2k−2j D(2k) = + ζ(2j), (12) 4k · (2k − 2)! (2k − 2j)! j=1 (−1)k+1(4k − 3)32k−1π2k Xk (−1)k+j+14j(3π)2k−2j D(2k) = + ζ(2j), (13) 2 · (2k)! (2k − 2j)! j=1 (−1)k+1(2k − 1)22k−2π2k kX−1 (−1)k+j+1(2π)2k−2j ζ(2k) = + ζ(2j), (14) (2k + 1)! (2k − 2j + 1)! j=1 (−1)k(4k + 1) ³π ´2k+1 Xk (−1)k+j ³π ´2k−2j+1 σ(2k + 1) = + ζ(2j), (15) 2 · (2k + 1)! 2 (2k − 2j + 1)! 2 j=1 (−1)k+1kπ2k kX−1 (−1)k+j+1π2k−2j ζ(2k) = + ζ(2j), (16) (2k + 1)! (2k − 2j + 1)! j=1 µ ¶ µ ¶ (−1)k+1(4k − 1)π2k+1 3 2k Xk (−1)k+j+1 3π 2k−2j+1 σ(2k + 1) = + ζ(2j), (17) 4 · (2k + 1)! 2 (2k − 2j + 1)! 2 j=1 (−1)k+1k(2π)2k kX−1 (−1)k+j+122k−2j+1π2k−2j ζ(2k) = + ζ(2j). (18) (2k + 2)! (2k − 2j + 2)! j=1 34 Suling Zhang and Chaoping Chen No. 3

Proof. Deﬁne the function f by π − x f(x) = , x ∈ (0, 2π). 2 Easy computation reveals the Fourier series of f on (0, 2π):

π − x X∞ sin nx = , x ∈ (0, 2π). (19) 2 n n=1 Integration term-by-term yields

x2 πx X∞ cos nx − + − ζ(2) = − , x ∈ [0, 2π]. 2 · 2! 2 n2 n=1 Clearly, if we integrate 2k − 1 times on each side of (19) from 0 to x, then we obtain

x2k πx2k−1 Xk (−1)jx2k−2j − + + ζ(2j) 2 · (2k)! 2 · (2k − 1)! (2k − 2j)! j=1 (20) X∞ cos nx = (−1)k , x ∈ [0, 2π]. n2k n=1 π Taking in (20) x = 2 and noticing that  nπ 0, n = 2m − 1; cos = 2 (−1)m, n = 2m, we conclude that

(−1)k+1(4k − 1)π2k Xk (−1)k+j+14jπ2k−2j D(2k) = + ζ(2j). (21) 2 · (2k)! (2k − 2j)! j=1 Taking x = π in (20) we conclude that

(−1)k+1π2k Xk (−1)k+j+1π2k−2j D(2k) = + ζ(2j). (22) 4k · (2k − 2)! (2k − 2j)! j=1

3π Taking x = 2 in (20) and noticing that X∞ cos 3nπ X∞ (−1)n−1 (−1)k+1 (−1)k 2 = (−1)k+1 = D(2k), n2k (2n)2k 22k n=1 n=1 we conclude that

(−1)k+1(4k − 3)32k−1π2k Xk (−1)k+j+14j(3π)2k−2j D(2k) = + ζ(2j). (23) 2 · (2k)! (2k − 2j)! j=1 Integrating on each side of (20) from 0 to 2π, we get

(−1)k+1(2k − 1)22k−2π2k kX−1 (−1)k+j+1(2π)2k−2j ζ(2k) = + ζ(2j). (24) (2k + 1)! (2k − 2j + 1)! j=1 Vol. 2 Recursion formulae for Riemann Zeta function and Dirchlet series 35

Integrating term-by-term on each side of (20) from 0 to x, we get

x2k+1 πx2k Xk (−1)jx2k−2j+1 − + + ζ(2j) 2 · (2k + 1)! 2 · (2k)! (2k − 2j + 1)! j=1 (25) X∞ sin nx = (−1)k , x ∈ [0, 2π]. n2k+1 n=1

π Taking x = 2 a in (25) nd noticing that  nπ (−1)m−1, n = 2m − 1; sin = 2 0, n = 2m,

we conclude that (−1)k(4k + 1) ³π ´2k+1 σ(2k + 1) = 2 · (2k + 1)! 2 (26) Xk (−1)k+j ³π ´2k−2j+1 + ζ(2j). (2k − 2j + 1)! 2 j=1

Taking x = π in (25) we conclude that

(−1)k+1kπ2k kX−1 (−1)k+j+1π2k−2j ζ(2k) = + ζ(2j). (27) (2k + 1)! (2k − 2j + 1)! j=1

3π Taking x = 2 in (25) and noticing that

X∞ sin 3nπ X∞ (−1)n−1 (−1)k 2 = (−1)k+1 = (−1)k+1σ(2k + 1), n2k+1 (2n − 1)2k+1 n=1 n=1 we conclude that µ ¶ (−1)k+1(4k − 1)π2k+1 3 2k σ(2k + 1) = 4 · (2k + 1)! 2 µ ¶ (28) Xk (−1)k+j+1 3π 2k−2j+1 + ζ(2j). (2k − 2j + 1)! 2 j=1

Integrating on each side of (25) from 0 to 2π, we get

(−1)k+1k(2π)2k kX−1 (−1)k+j+122k−2j+1π2k−2j ζ(2k) = + ζ(2j). (29) (2k + 2)! (2k − 2j + 2)! j=1

The proof of Theorem 1 is complete. Remark Form recurrence formula (14), or (16), or (18) we can obtain values of ζ(2k), for examples,

π2 π4 π6 π8 π10 ζ(2) = , ζ(4) = , ζ(6) = , ζ(8) = , ζ(10) = . 6 90 945 9450 93555 36 Suling Zhang and Chaoping Chen No. 3

By using values of ζ(2k), we can conclude values of D(2k) from (11), or (12), or (13), for examples,

X∞ (−1)n−1 π2 D(2) = = , n2 12 n=1 X∞ (−1)n−1 7π4 D(4) = = , n4 720 n=1 X∞ (−1)n−1 31π6 D(6) = = . n6 30240 n=1

By using values of ζ(2k), we can also conclude values of σ(2k) from (15), or (17), for examples,

X∞ (−1)n−1 π σ(1) = = , 2n − 1 4 n=1 X∞ (−1)n−1 π3 σ(3) = = , (2n − 1)3 32 n=1 X∞ (−1)n−1 5π5 σ(5) = = . (2n − 1)5 1536 n=1

Theorem 2. For k ∈ N, we have   1 (−1)k+1π2k kX−1 (−1)k+j+14jπ2k−2j D(2k) =  + + D(2j) , (30) 4k − 1 2 · (2k)! (2k − 2j)! j=1 (−1)k+1π2k kX−1 (−1)k+j+1π2k−2j D(2k) = + D(2j), (31) 2 · (2k + 1)! (2k − 2j + 1)! j=1 (−1)k ³π ´2k+1 Xk (−1)k+j ³π ´2k−2j+1 σ(2k + 1) = + D(2j), (32) 2 · (2k + 1)! 2 (2k − 2j + 1)! 2 j=1 (−1)k+1π2k Xk (−1)k+j+1π2k−2j ζ(2k) = + D(2j). (33) 2 · (2k)! (2k − 2j)! j=1

Proof. Deﬁne the function g by

g(x) = x, x ∈ (−π, π).

Easy computation reveals the Fourier series of f on (−π, π):

X∞ (−1)n−1 sin nx x = 2 , x ∈ (−π, π). (34) n n=1

Integration term-by-term yields

x2 X∞ (−1)n−1 cos nx − D(2) = − , x ∈ [−π, π]. 2 · 2! n2 n=1 Vol. 2 Recursion formulae for Riemann Zeta function and Dirchlet series 37

Clearly, if we integrate 2k − 1 times on each side of (34) from 0 to x, then we obtain

x2k Xk (−1)jx2k−2j X∞ (−1)n−1 cos nx + D(2j) = (−1)k , x ∈ [−π, π]. (35) 2 · (2k)! (2k − 2j)! n2k j=1 n=1

Taking x = π/2 in (35) we conclude that   1 (−1)k+1π2k kX−1 (−1)k+j+14jπ2k−2j D(2k) =  + + D(2j) . (36) 4k − 1 2 · (2k)! (2k − 2j)! j=1

Integrating on each side of (35) from 0 to π, we conclude that

(−1)k+1π2k kX−1 (−1)k+j+1π2k−2j D(2k) = + D(2j). (37) 2 · (2k + 1)! (2k − 2j + 1)! j=1

Integrating term-by-term on each side of (35) from 0 to x, we get

x2k+1 Xk (−1)jx2k−2j+1 + D(2j) 2 · (2k + 1)! (2k − 2j + 1)! j=1 (38) X∞ (−1)n+1 sin nx = (−1)k , x ∈ [−π, π]. n2k+1 n=1

Taking x = π/2 in (38) we conclude that

(−1)k ³π ´2k+1 Xk (−1)k+j ³π ´2k−2j+1 σ(2k + 1) = + D(2j). (39) 2 · (2k + 1)! 2 (2k − 2j + 1)! 2 j=1

Integrating on each side of (38) from 0 to π, we get

(−1)k+1π2k Xk (−1)k+j+1π2k−2j ζ(2k) = + D(2j). (40) 2 · (2k)! (2k − 2j)! j=1

The proof of Theorem 2 is complete. Theorem 3. For k ∈ N, we have

(−1)k ³π ´2k+1 Xk (−1)k+j ³π ´2k−2j+1 σ(2k + 1) = + δ(2j), (41) 2 · (2k)! 2 (2k − 2j + 1)! 2 j=1 (−1)k+1π2k kX−1 (−1)k+j+1π2k−2j δ(2k) = + δ(2j), (42) 8 · (2k − 1)! 2 · (2k − 2j)! j=1 (−1)k ³π ´2k kX−1 (−1)k+j ³π ´2k−2j δ(2k) = + δ(2j), (43) 2 · (2k − 1)! 2 (2k − 2j)! 2 j=1 (−1)k+1π2k kX−1 (−1)k+j+1π2k−2j δ(2k) = + δ(2j). (44) 4 · (2k)! (2k − 2j + 1)! j=1 38 Suling Zhang and Chaoping Chen No. 3

Proof. Deﬁne the function h by

h(x) = |x|, x ∈ [−π, π].

Easy computation reveals the Fourier series of f on [−π, π]:

π 4 X∞ cos(2n − 1)x |x| = − , x ∈ [−π, π]. (45) 2 π (2n − 1)2 n=1 Taking in (45) x = 0, we get

X∞ 1 π2 δ(2) = = . (2n − 1)2 8 n=1 Thus we have from (45)

π X∞ cos(2n − 1)x x − δ(2) = − , x ∈ [0, π]. (46) 4 (2n − 1)2 n=1 Integrating term-by-term, we have from (46)

π x2 X∞ sin(2n − 1)x · − δ(2)x = − , x ∈ [0, π]. 4 2! (2n − 1)3 n=1 Clearly, if we integrate 2k − 1 times on each side of (46) from 0 to x, then we obtain

π x2k Xk (−1)jx2k−2j+1 X∞ sin(2n − 1)x · + δ(2j) = (−1)k , x ∈ [0, π]. (47) 4 (2k)! (2k − 2j + 1)! (2n − 1)2k+1 j=1 n=1 Taking in (47) x = π/2 we conclude that

(−1)k ³π ´2k+1 Xk (−1)k+j ³π ´2k−2j+1 σ(2k + 1) = + δ(2j). (48) 2 · (2k)! 2 (2k − 2j + 1)! 2 j=1 Integrating on each side of (47) from 0 to π, we conclude that

(−1)k+1π2k kX−1 (−1)k+j+1π2k−2j δ(2k) = + δ(2j). (49) 8 · (2k − 1)! 2 · (2k − 2j)! j=1 Integrating term-by-term on each side of (47) from 0 to x, we obtain that

π x2k+1 kX+1 (−1)jx2k−2j+2 · + δ(2j) 4 (2k + 1)! (2k − 2j + 2)! j=1 (50) X∞ cos(2n − 1)x = (−1)k+1 , x ∈ [0, π]. (2n − 1)2k+2 n=1 Taking x = π/2 in (50), we conclude that

(−1)k ³π ´2k kX−1 (−1)k+j ³π ´2k−2j δ(2k) = + δ(2j). (51) 2 · (2k − 1)! 2 (2k − 2j)! 2 j=1 Vol. 2 Recursion formulae for Riemann Zeta function and Dirchlet series 39

Integrating on each side of (50) from 0 to π, we conclude that

(−1)k+1π2k kX−1 (−1)k+j+1π2k−2j δ(2k) = + δ(2j). (52) 4 · (2k)! (2k − 2j + 1)! j=1

The proof of Theorem 3 is complete. Remark. Form recurrence formula (42), (43) or (44), we can obtain values of δ(2k), for examples,

X∞ 1 π2 δ(2) = = , (2n − 1)2 8 n=1 X∞ 1 π4 δ(4) = = , (2n − 1)4 96 n=1 X∞ 1 π6 δ(6) = = . (2n − 1)6 960 n=1

References

[1] Alfred van der Poorten, A proof missed by Euler, ShùxuéY`ıl´ın(Translations of Math- ematics) 2(1980), pp. 47-63. (Chinese) [2] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill, 1978. [3] E. Grosswald, Comments on Some formulae of Ramnujan, Acta. Arith. 21(1972), pp. 25-34. [4] E. Grosswald, Die Werte der Riemannschen Zeta Function an ungeraden Argumen- stellen, GöttingenNachrichten, 1970, pp. 9-13. [5] Group of Compilation, ShùxuéShˇoucè,Handbook of Mathematics, Higher Education Press, Beijing, China, 1990. (Chinese). [6] G. H. Hardy, A formula of Ramanujan, J. London Math. Soc. 3 (1928), pp. 238-240. [7] L.-G. Hua, G¯aodˇengShùxuéYˇinlùn,Introduction to Advanced Mathematics, Vol.1, Fascicule 2, Science Press, Beijing, China, 1979. pp. 281-283 (Chinese). [8] J.-Ch. Kuang, ChángyòngBùdˇengsh`ı,Applied Inequalities, 2nd edition, Hunan Edu- cation Press, Changsha City, Hunan Province, China, 1993 (Chinese). P +∞ 1 ¯ [9] J.-F. Lin, Elementary solvabilities of n=1 n2m , Shùxuéde Sh´ıjiànyùRènshi, Math. Practice Theory, 29(1999), No.3, pp. 14-18 (Chinese). [10] G.-D. Liu, A class of evaluating formulae including Riemann Zeta, K¯exuéT¯ongbào, Chinese Sci. Bull., 44(1999), No.2, pp. 146-148 (Chinese). [11] H. M. Srivastava, Certain families of rapidly convergent series representations for ζ(2n + 1), Math. Sci. Res. Hot-Line 1(1997), No.6, pp. 1-6. [12] H. M. Srivastava, Further series representations for ζ(2n + 1), Appl. Math. Comput. 97(1998), pp. 1-15. [13] H. M. Srivastava, Some rapidly converging series for ζ(2n + 1), Proc. Amer. Math. Soc. 127(1999), pp. 385-396. 40 Suling Zhang and Chaoping Chen No. 3

[14] H. M. Srivastava, Some simple algorithms for the evaluations and representations of the Riemann Zeta function at positive integer arguments, J. Math. Anal. Appl. 246(2000), pp. 331-351. [15] H. M. Srivastava and J. Choi, Series Associated with the Zeta and Related Functions, Kluwer Academic Publishers, Dordrecht, Boston, and London, 2001. [16] H. M. Srivastava, M. L. Glasser, and V. S. Adamchik, Some definite integrals associated with the Riemann Zeta function, Z. Anal. Anwendungen 19(2000), pp. 831-846. [17] H. M. Srivastava and H. Tsumura, A certain class of rapidly convergent series representations for ζ(2n + 1), J. Comput. Appl. Math. 118(2000), pp. 323-335. [18] Zh.-X. Wang and D.-R. Guo, Tèsh¯uHánshùGàilùnIntroduction to Special Function, The Series of Advanced Physics of Peking University, Peking University Press, Beijing, China, 2000 (Chinese). [19] L.-Zh. Xu (L. C. Hsu) and X.-H. Wang, ShùxuéF¯enx¯i Zh¯ongde F¯angfˇahéL`ıtˇi Xuˇanjiˇang,Methods of Mathematical Analysis and Selected Examples, Revised Edition, Higher Education Press, Beijing, China, 1983. pp. 281-283 (Chinese). [20] N.-Y. Zhang, Values of Riemannjan formula and Riemann Zeta function at positive odd numbers, ShùxuéJ`ınzhˇan,Adv. Math.(China) 12(1983), No.1, pp. 61-71 (Chinese). [21] Q.-T. Zhuang and N.-Y. Zhang, FùbiànHánshù,Complex Functions, Beijing Univer- sity Press, Beijing, China, 1984. pp.10 and pp. 249 (Chinese). [22] W.-P. Zhang, On identities of Riemann Zeta function, K¯exuéT¯ongbào,Chinese Sci. Bull. 36(1991), No.4, pp. 250-253 (Chinese). Scientia Magna Vol. 2 (2006), No. 3, 41-43

Smarandache semiquasi near-rings1

Arun S. Muktibodh Mohota Science College, Umred Rd. Nagpur-440009, India

Abstract G. Pilz [1] has deﬁned near-rings and semi-near-rings. In this paper we introduce the concepts of quasi-near ring and semiquasi-near ring. We have also deﬁned Smarandache semiquasi-near-ring. Some examples are constructed. We have posed some open problems. Keywords Near-ring, semi-near-ring, quasi-near-ring, semiquasi-near-ring, Smarandache semiquasi-near-ring.

§1. Introduction

In the paper [2] W.B. Kandasamy has introduced a new concept of Smarandache semi- near ring. These are associative rings. We have defined a new concepts of quasi-near ring and Smarandache semiquasi-near-ring. These are non associative rings. Definition 1.1. An algebraic structure (Q, +, ·) is called a quasi-near-ring (or a right quasi-near-ring) if it satisfies the following three conditions: 1.(Q, +) is a group (not necessarily abelian). 2.(Q, ·) is a quasigroup.

3.(n1 + n2) · n3 = n1 · n3 + n2 · n3 for all n1, n2, n3 ∈ Q (right distributive law). Example 1.1. Let Q = {1, 2, 3, 4} and the two binary operations are deﬁned on Q by the following tables;

+ 1 2 3 4 . 1 2 3 4 1 1 2 3 4 1 4 2 1 3 2 2 3 4 1 2 1 3 4 2 3 3 4 1 2 3 3 1 2 4 4 4 1 2 3 4 2 4 3 1

Deﬁnition 1.2. An algebraic system (S, +, ·) is called a semiquasi-near-ring (or right semiquasi-near-ring) if it satisﬁes the following three conditions: 1. (S, +) is a quasigroup (not necessarily abelian). 2. (S, ·) is a quasigroup.

3. (n1 + n2) · n3 = n1 · n3 + n2 · n3 for all n1, n2, n3 ∈ S (right distributive law). Example 1.2. Consider the algebraic system (S, +, ·) where S = {1, 2, 3, 4} deﬁned by the following tables; 1This work is supported by UGC under Minor project F. No. 23-245/06. 42 Arun S. Muktibodh No. 3

+ 1 2 3 4 . 1 2 3 4 1 1 3 4 2 1 4 2 1 3 2 4 2 1 3 2 1 3 4 2 3 2 4 3 1 3 3 1 2 4 4 3 1 2 4 4 2 4 3 1 Example 1.3. We know that integers Z with subtraction ( - ) forms a quasigroup. (Z, ·) is a quasigroup and subtraction of integers distributes over multiplication. Thus (Z, −, ·) is a semiquasi-near-ring. Deﬁnition 1.3 We know that integers Z with subtraction ( - ) forms a quasigroup. (Z, ·) is a quasigroup and subtraction of integers distributes over multiplication. Thus (Z, −, ·) is a semiquasi-near-ring. Example 1.4. Consider the semiquasi-near-ring (S, +, ·) deﬁned by the following tables;

+ 1 2 3 4 . 1 2 3 4 1 1 3 4 2 1 4 3 1 2 2 4 2 1 3 2 1 2 4 3 3 2 4 3 1 3 3 4 2 1 4 3 1 2 4 4 2 1 3 4 one can easily verify that addition distributes over multiplication from right as well as S contains N = {4} properly which is a quasi-near-ring. Thus (S, +, ·) is a Smarandache semiquasi-near-ring. Example 1.5. Let R be the set of reals. We know that (R, +) is a group and hence a quasigroup. Also, R w.r.t. division is a quasigroup, that is (R, ÷) is a quasigroup. More over, addition distributes over division from right. Thus (R, +, ÷) is a semiquasi-near-ring. Let Q be the set of non-zero rationals. Then (Q, +) is a group. Also, (Q, ÷) is a quasigroup. Addition distributes over division. Hence (Q, +, ÷) is a quasi-near-ring. We know that R ⊃ Q. Therefore, (R, +, ÷) is a Smarandache semiquasi-near-ring. We now show by an example that there do exist semiquasi-near -rings which are not Smarandache semiquasi-near-rings. Consider example 1.2 where we can not have a quasi-near-ring contained in S. We give below the example of a smallest Smarandache semiquasi-near-ring which is not a near- ring. Example 1.6. Consider the semiquasi-near ring (S = {1, 2, 3}, +, ·) deﬁned by the following tables;

+ 1 2 3 . 1 2 3 1 1 3 2 1 3 2 1 2 3 2 1 2 1 3 2 3 2 1 3 3 2 1 3 Vol. 2 Smarandache semiquasi near-rings 43

One can easily verify that (S = {1, 2, 3}, +, ·) is a semiquasi-near-ring. Moreover, N = {3} ⊂ S and (N, +, ·) is a quasi-near ring. Therefore (S = {1, 2, 3}, +, ·) is a Smarandache semiquasi- near-ring. We now show by an example that there do exist semiquasi-near -rings which are not Smarandache semiquasi-near-rings. Consider example 1.2 where we can not have a quasi-near-ring contained in S. We give below the example of a smallest Smarandache semiquasi-near-ring which is not a near- ring. Definition 1.4. N is said to be an Anti-Smarandache semiquasi-near-ring if N is a quasi- near-ring and has a proper subset A such that A is a semiquasi-near-ring under the same operations as of N. Example 1.7. In example 1.5 (R, +, ÷) is also a quasi-near-ring which contains a semiquasi- near-ring (Q, +, ÷). Thus we can say that (R, +, ÷) is an Anti-Smarandache semiquasi-near-ring. We propose the following: Problem 1. Do there exist a finite Smarandache semiquasi-near-ring such that the order of the quasi-near-ring contained in it is greater than 1 ? Problem 2. How to construct finite Anti-Smarandache semiquasi-near-rings ?

References

[1] G. Pilz, Near-rings, North-Holland Publ. and Co., 1977. [2] W.B.Kandasamy, Smarandache near-rings and their generalizations, Smarandache No- tions, book series, Amerisan Research Press,14(2004), pp. 315-319. [3] R.Padilla, Smarandache Algebraic structures, Buletin of Pure and Applied Sciences, Delhi, 1(1998), pp. 119- 121. http://www.gallup.unm.edu/ smarandache/ALG-S-TXT.TXT [4] Ashbacher, Charles, On Numbers that are Pseudo-Smarandache and Smarandache perfect, Smarandache Notions Journal, 14(2004), pp. 40 41. [5] Robinson Derek J.S., A course in the theory of Groups, Springer verlag, 1996. Scientia Magna Vol. 2 (2006), No. 3, 44-47

On exponentially harmonic numbers

J´ozsefS´andor Department of Mathematics and Computer Sciences, Babe¸s-Bolyai University of Cluj, Cluj, Romania

Abstract In this note there are introduced two new concepts of e-harmonic numbers, and a new concept of e-perfect number. Their initial study is provided. Keywords Exponential divisors, harmonic numbers, e-perfect numbers.

§1. Introduction

Let σ(n) and d(n) denote the sum, resp. number of divisors of n. In 1948 O. Ore [12] called a number n harmonic if σ(n)|nd(n) (1)

See e.g. G. L. Cohen and R. M. Sorli [2] for such numbers. If σk(n) denotes the sum of k-th powers of divisors of n (k ≥ 1 integer), then G. L. Cohen and D. Moujie [3] introduced k-harmonic numbers by

σk(n)|nd(n) (2)

A perfect number is always harmonic (see [12]), and Ore conjectured that all harmonic numbers are even. This is a quite deep conjecture, since, if true, clearly would imply the non-existence of odd perfect numbers. ¡ ¢ n ∗ ∗ A divisor d of n is called unitary divisor if d, d = 1. If σ (n), d (n) are the sum, resp. number of unitary divisors of n, then n is called unitary harmonic if

σ∗(n)|nd∗(n), (3)

see K. Nageswara Rao [11], P. Hagis and G. Lord [5], Ch. Wall [19]. A divisor d of n is called a bi-unitary divisor, if the greatest common unitary divisor of d n ∗∗ ∗∗ and d is 1. If σ (n), d (n) are the sum, and number of bi-unitary divisors of n, recently we have introduced (see [16]) bi-unitary harmonic numbers by

σ∗∗(n)|nd∗∗(n) (4)

For inﬁnitary harmonic numbers, related to the concept of an inﬁnitary divisor, see P. Hagis and G. L. Cohen [7].

a1 ar Let n > 1 be a positive integer having the prime factorization n = p1 . . . pr . A divisor b1 br d of n is called exponential divisor, if d = p1 . . . pr where b1|a1, . . . , br|ar. This notion is due to E. G. Straus and M. V. Subbarao [17]. Let σe(n) and de(n) be the sum and number of, Vol. 2 On exponentially harmonic numbers 45

exponential divisors of n. Let by convention σe(1) = de(1) = 1. Straus and Subbarao have introduced e-perfect numbers n by

σe(n) = 2n. (5) They proved the non-existence of odd e-perfect numbers, with related other results. For results on e-superperfect numbers (i.e. satisfying σe(σe(n)) = 2n), see [8]. For density problems, e-perfect numbers not divisible by 3, or e-multiperfect numbers, see P. Hagis [6], L. Lucht [10],

J. Fabrykowski and M. V. Subbarao [4], W. Aiello et al. [1]. For results on de(n), we quote J.

M. DeKoninck and A. Ivić[9]. For the exponential totient function ϕe(n), see J. Sándor[13]. For e-convolution and a survey connected to the Möbiusfunction, see J. Sándorand A. Bege [14]. For multiplicatively e-perfect numbers, see J. Sándor[15].

§2. Exponential harmonic numbers

The aim of this note is study two notions of e-harmonic numbers. An integer n will be called e-harmonic of type 1 if

σe(n)|nde(n) (6) In all examples of section 1 the harmonic numbers notions were suggested by the consideration of the harmonic means of the considered divisors. For example, if 1 = d1 < d2 < ··· < dr = n are all divisors of n, then their harmonic mean is µ ¶ 1 1 H(n) = r/ + ··· + . d1 dr Since X 1 1 X n 1 X σ(n) = = dr = , dr n dr n n we get nd(n) H(n) = , (7) σ(n) so a harmonic number n is a number such that H(n) is an integer. E.g. for H(n) = 2 we get the so-called “balanced numbers”proposed by M. V. Subbarao [18], with single solution n = 6. (e) (e) Now, if we consider the harmonic mean He(n) of the exponential divisors d1 , . . . , dr , then r He(n) = P 1 , (e) di a where r = de(n) denotes the number of exponential (or e-) divisors of n. Let p be a prime power. Then the e-divisors of pa are pd with d|a, so X 1 X 1 1 X = = pa−d (e) pa pa di d|a d|a in this case. When n = paqb (p 6= q primes) one obtains similarly     X 1 X 1 X 1 X 1 = =     d(e) pd1 qd2 pd1 qd2 i d1|a,d2|b d1|a d2|b 46 J´ozsefS´andor No. 3     1 X X =  pa−d1   qb−d2  . paqb d1|a d2|b

a1 a2 ar In the general case, when n = p1 p2 . . . pr , one has

nde(n) He(n) = , (8) Se(n) where   Yr X  ai−di  Se(n) = pi (9)

i=1 di|ai

We say that n is e-harmonic of type 2 if He(n) is integer, i.e.

Se(n)|nde(n), (10)

where Se(n) is given by (9). Theorem 1. If n is squarefree, then it is e-harmonic of both types. Proof. If n is squarefree, i.e. n = p1p2 . . . pr, then clearly by deﬁnitions of σe(n) and Yr X  1−di  Se(n) one has σe(n) = p1p2 . . . pr = n and Se(n) = pi = 1, so (6) and (8) are

i=1 di|1 satisﬁed. We shall see later (see the Remark after Theorem 3), that there exist also numbers with this property, which are not squarefree.

a1 ar Theorem 2. Let n = p1 . . . pr be the prime factorization of n > 1. If n is e-perfect, then n is e-harmonic of type 1 if and only if at least one of a1, . . . , ar is not a perfect square.

Proof. If σe(n) = 2n, then (6) gives 2|de(n). It is well-known that de(n) = d(a1) . . . d(ar),

so at least one of d(a1), . . . , d(ar) must be even. But, it is well-known that d(a) is even iﬀ a is not a perfect square, so the result follows. Remark. 1) Since e.g. 22 ·33 ·52, 23 ·32 ·52, 24 ·33 ·52 ·112, 26 ·33 ·52 ·72 ·132, 27 ·32 ·52 ·72 ·132 are e-perfect numbers, by the above theorem, these are also e-harmonic numbers of type 1.

2) Similarly, if n is e − k perfect, i.e. σe(n) = kn (k ≥ 2 integer), see [?], then n is e-harmonic of type 1 iﬀ

k|d(a1)d(a2) . . . d(ar) (11) In what follows we shall introduce another new notion. We say that n is modiﬁed e- perfect number, if

Se(n)|n, (12) 2 3 2 where Se(n) is given by (9). Since Se(p) = 1, Se(p ) = p + 1, Se(p ) = p + 1, we have 2 2 2 2 3 2 2 2 2 Se(2 · 3 ) = (2 + 1)(3 + 1) = 2 · 3, Se(2 · 3 · 5 ) = (2 + 1)(3 + 1)(5 + 1) = 2 · 3 · 5, 3 2 2 2 3 2 2 2 3 2 3 2 2 Se(2 · 3 · 5 ) = (2 + 1)(3 + 1)(5 + 1) = 2 · 3 · 5, so 2 · 3 , 2 · 3 · 5 , 2 · 3 · 5 (which are also e-perfect) are modified e-perfect numbers. Clearly n = 24 · 32 · 112 (which is e-perfect) is 4 2 3 4 2 3 not modified e-perfect, since by Se(p ) = 1 + p + p , we have Se(2 ) = 1 + 2 + 2 = 13. Theorem 3. If n is modified e-perfect, then it is harmonic of type 2. Proof. This follows at once from (12) and (8). Remark. Thus 22 · 33 · 52 and 23 · 32 · 52 are e-harmonic numbers of both types (though they are not squarefree). Vol. 2 On exponentially harmonic numbers 47 References

[1] W. Aiello et al., On the existence of e-multiperfect numbers, Fib. Quart. 25(1987), pp. 65-71. [2] G. L. Cohen and R. M. Sorli, Harmonic seeds, Fib. Quart., 36(1998), pp. 386-390. [3] G. L. Cohen and D. Moujie, On a generalization of Ore’s harmonic numbers, Nieuw Arch. Wiskunde, 16(1998), No.3, pp. 161-172. [4] J. M. DeKoninck and A. Ivić,An asymptotic formula for reciprocals of logarithms of certain multiplicative functions, Canad. Math. Bull., 21(1978), pp. 409-413. [5] J. Fabrykowski and M. V. Subbarao, On e-perfect numbers not divisible by 3, Nieuw Arch. Wiskunde (4), 4(1986), pp. 165-173. [6] P. Hagis and G. Lord, Unitary harmonic numbers, Proc. Amer. Math. Soc., 51(1975), pp. 1-7. [7] P. Hagis, Some results concerning exponential divisors, Int. J. Math. Math. Sci., 11(1988), pp. 3434-349. [8] P. Hagis and G. L.Cohen, Infinitely harmonic numbers, Bull. Austral. Math. Soc., 41(1990), pp. 151-158. [9] J. Hanumanthachari et al., On e-perfect numbers, Math. Student, 46(1978), pp. 71-80. [10] L. Lucht, On the sum of exponential divisors and its iterates, Arch. Math., Basel, 27(1976), pp. 383-386. [11] K.Nageswara Rao, On some unitary divisor functions, Scripta Math., 28(1967), pp. 347-351. [12] O. Ore, On the averages of the divisors of a number, Amer. Math. Monthly, 55(1948), pp. 615-619. [13] J. Sándor,On an exponential totient function, Studia Univ. Babe¸s-Bolyai Math. 41(1996), pp. 91-94. [14] J. Sándorand A. Bege, The Möbiusfunction: generalizations and extensions, Adv. Stud. Contemp. Math., 6(2003), No.2, pp. 77-128. [15] J. Sándor,On multiplicatively e-perfect numbers, (to appear). [16] J. Sándor,Bi-unitary harmonic numbers, (to appear). [17] E. G. Straus and M. V. Subbarao, On exponential divisors, Duke Math. J. 41(1974), pp. 465-471. [18] M. V. Subbarao, Problem E1558, Amer. Math. Monthly, 70(1963), pp92, Solution in 70(1963), pp. 1009-1010. [19] Ch. R. Wall, Unitary harmonic numbers, Fib. Quart., 21(1983), pp. 18-25. Scientia Magna Vol. 2 (2006), No. 3, 48-53

Parastrophic invariance of Smarandache quasigroups

Tèm´ıtópéGbóláhànJa´ıyéo.lá 1 Department of Mathematics, Obafemi Awolowo University, Ile Ife, Nigeria

Abstract Every quasigroup (L, ·) belongs to a set of 6 quasigroups, called parastrophes

denoted by (L, πi), i ∈ {1, 2, 3, 4, 5, 6}. It is shown that (L, πi) is a Smarandache quasigroup

with associative subquasigroup (S, πi) ∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if for any of some four

j ∈ {1, 2, 3, 4, 5, 6},(S, πj ) is an isotope of (S, πi) or (S, πk) for one k ∈ {1, 2, 3, 4, 5, 6} such

that i 6= j 6= k. Hence, (L, πi) is a Smarandache quasigroup with associative subquasigroup

(S, πi) ∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if any of the six Khalil conditions is true for any of

some four of (S, πi). Keywords Parastrophes, Smarandache quasigroups, isotopic.

§1. Introduction

The study of the Smarandache concept in groupoids was initiated by W.B. Vasantha Kan- dasamy in [18]. In her book [16] and first paper [17] on Smarandache concept in loops, she defined a Smarandache loop as a loop with at least a subloop which forms a subgroup under the binary operation of the loop. Here, the study of Smarandache quasigroups is continued after their introduction in Muktibodh [9] and [10]. Let L be a non-empty set. Define a binary operation (·) on L : if x · y ∈ L ∀ x, y ∈ L,(L, ·) is called a groupoid. If the system of equations ; a · x = b and y · a = b have unique solutions for x and y respectively, then (L, ·) is called a quasigroup. Furthermore, if ∃ a ! element e ∈ L called the identity element such that ∀ x ∈ L, x · e = e · x = x,(L, ·) is called a loop. It can thus be seen clearly that quasigroups lie in between groupoids and loops. So, the Smarandache concept needed to be introduced into them and studied since it has been introduced and studied in groupoids and loops. Definitely, results of the Smarandache concept in groupoids will be true in quasigroup that are Smarandache and these together will be true in Smarandache loops. It has been noted that every quasigroup (L, ·) belongs to a set of 6 quasigroups, called adjugates by Fisher and Yates [6], conjugates by Stein [15], [14] and Belousov [2] and parastrophes by Sade [12]. They have been studied by Artzy [1] and a detailed study on them can be found in [11], [4] and [5]. So for a quasigroup (L, ·), its parastrophes are denoted by (L, πi), i ∈ {1, 2, 3, 4, 5, 6} hence one can take (L, ·) = (L, π1). For more on quasigroup, loops and their properties, readers should check [11], [3], [4], [5], [7] and [16]. Let (G, ⊕) and (H, ⊗) be two distinct quasigroups. The triple (A, B, C) such that A, B, C :(G, ⊕) → (H, ⊗) are bijections

1On Doctorate Programme at the University of Agriculture Abeokuta, Nigeria. Vol. 2 Parastrophic invariance of Smarandache quasigroups 49

is said to be an isotopism if and only if xA ⊗ yB = (x ⊕ y)C ∀ x, y ∈ G. Thus, H is called an isotope of G and they are said to be isotopic.

In this paper, it will be shown that (L, πi) is a Smarandache quasigroup with associative

subquasigroup (S, πi) ∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if for any of some four j ∈ {1, 2, 3, 4, 5, 6},

(S, πj) is an isotope of (S, πi) or (S, πk) for one k ∈ {1, 2, 3, 4, 5, 6} such that i 6= j 6= k. Hence,

it can be concluded that (L, πi) is a Smarandache quasigroup with associative subquasigroup

(S, πi) ∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if any of the six Khalil conditions is true for any of some

four of (S, πi).

§2. Preliminaries

Definition 2.1 Let (L, ·) be a quasigroup. If there exists at least a non-trivial subset S ⊂ L such that (S, ·) is an associative subquasigroup in L, then L is called a Smarandache quasigroup (SQ). Definition 2.2 Let (L, θ) be a quasigroup. The 5 parastrophes or conjugates or adjugates of (L, θ) are quasigroups whose binary operations θ∗ , θ−1 , −1θ , (θ−1)∗ , (−1θ)∗ defined on L are given by : (a) (L, θ∗): yθ∗x = z ⇔ xθy = z ∀ x, y, z ∈ L. (b) (L, θ−1): xθ−1z = y ⇔ xθy = z ∀ x, y, z ∈ L. (c) (L, −1θ): z −1θy = x ⇔ xθy = z ∀ x, y, z ∈ L. (d) (L, (θ−1)∗): z(θ−1)∗x = y ⇔ xθy = z ∀ x, y, z ∈ L. (e) (L, (−1θ)∗): y(−1θ)∗z = x ⇔ xθy = z ∀ x, y, z ∈ L. Definition 2.3 Let (L, θ) be a loop.

(a) Rx and Lx represent the left and right translation maps in (L, θ) ∀ x ∈ L. ∗ ∗ ∗ (b) Rx and Lx represent the left and right translation maps in (L, θ ) ∀ x ∈ L. −1 (c) Rx and Lx represent the left and right translation maps in (L, θ ) ∀ x ∈ L. −1 (d) Rx and L represent the left and right translation maps in (L, θ) ∀ x ∈ L. ∗ ∗ −1 ∗ (e) Rx and Lx represent the left and right translation maps in (L, (θ ) ) ∀ x ∈ L. ∗ ∗ −1 ∗ (f) Rx and L represent the left and right translation maps in (L, ( θ) ) ∀ x ∈ L. Remark 2.1 If (L, θ) is a loop, (L, θ∗) is also a loop(and vice versa) while the other adjugates are quasigroups. Lemma 2.1 If (L, θ) is a quasigroup, then ∗ ∗ −1 −1 ∗ −1 ∗ −1 1. Rx = Lx ,Lx = Rx , Lx = Lx , Rx = Rx , Rx = Lx , Lx = Rx ∀ x ∈ L. ∗−1 ∗−1 ∗ ∗−1 ∗ ∗−1 2. Lx = Rx , Rx = Lx , Rx = Rx = Lx , Lx = Lx = Rx ∀ x ∈ L. Proof. The proof of these follows by using Definition 2.2 and Definition 2.3. ∗ ∗ ∗ ∗ ∗ (1) yθ x = z ⇔ xθy = z ⇒ yθ x = xθy ⇒ yRx = yLx ⇒ Rx = Lx. Also, yθ x = xθy ⇒ ∗ ∗ xLy = xRy ⇒ Ly = Ry. −1 −1 xθ z = y ⇔ xθy = z ⇒ xθ(xθ z) = z ⇒ xθzLx = z ⇒ zLxLx = z ⇒ LxLx = I. Also, −1 −1 −1 xθ (xθy) = y ⇒ xθ yLx = y ⇒ yLxLx = y ⇒ LxLx = I. Hence, Lx = Lx ∀ x ∈ L. −1 −1 −1 z( θ)y = x ⇔ xθy = z ⇒ (xθy)( θ)y = x ⇒ xRy( θ)y = x ⇒ xRyRy = x ⇒ RyRy = I. −1 −1 Also, (z( θ)y)θy = z ⇒ zRyθy = z ⇒ zRyRy = z ⇒ RyRy = I. Thence, Ry = Ry ∀ x ∈ L. −1 ∗ −1 ∗ ∗ ∗ ∗ z(θ ) x = y ⇔ xθy = z, so, xθ(z(θ ) x) = z ⇒ xθzRx = z ⇒ zRxLx = z ⇒ RxLx = I. 50 Tèm´ıtópéGbóláhànJa´ıyéo.lá No. 3

−1 ∗ −1 ∗ ∗ ∗ ∗ −1 Also, (xθy)(θ ) x = y ⇒ yLx(θ ) x = y ⇒ yLxRx = y ⇒ LyRx = I. Whence, Rx = Lx . −1 ∗ −1 ∗ −1 ∗ ∗ ∗ y( θ) z = x ⇔ xθy = z, so, y( θ) (xθy) = x ⇒ y( θ) xRy = x ⇒ xRyLy = x ⇒ RyLy = −1 ∗ ∗ ∗ ∗ ∗ −1 I. Also, (y( θ) z)θy = z ⇒ zLyθy = z ⇒ zLyRy = z ⇒ LyRy = I. Thus, Ly = Ry . (2) These ones follow from (1). Lemma 2.2 Every quasigroup which is a Smarandache quasigroup has at least a subgroup. Proof. If a quasigroup (L, ·) is a SQ, then there exists a subquasigroup S ⊂ L such that (S, ·) is associative. According [8], every quasigroup satisfying the associativity law has an identity hence it is a group. So, S is a subgroup of L. Theorem 2.1 (Khalil Conditions [13]) A quasigroup is an isotope of a group if and only if any one of some six identities are true in the quasigroup.

§3. Main Results

Theorem 3.1 (L, θ) is a Smarandache quasigroup of with associative subquasigroup (S, θ) if and only if any of the following equivalent statements is true. 1. (S, θ) is isotopic to (S, (θ−1)∗). 2. (S, θ∗) is isotopic to (S, θ−1). 3. (S, θ)is isotopic to (S, (−1θ)∗). 4. (S, θ∗) is isotopic to (S, −1θ).

Proof. L is a SQ with associative subquasigroup S if and only if s1θ(s2θs3) = (s1θs2)θs3 ⇔

Rs2 Rs3 = Rs2θs3 ⇔ Ls1θs2 = Ls2 Ls1 ∀ s1, s2, s3 ∈ S. The proof of the equivalence of (1) and (2) is as follows. L = L L ⇔ L−1 = s1θs2 s2 s1 s1θs2 −1 −1 −1 −1 −1 −1 Ls2 Ls1 ⇔ Ls1θs2 = Ls1 Ls2 ⇔ (s1θs2)θ s3 = s2θ (s1θ s3) ⇔ (s1θs2)Rs3 = s2θ s1Rs3 = −1 ∗ −1 ∗ ∗ −1 s1Rs3 (θ ) s2 ⇔ (s1θs2)Rs3 = s1Rs3 (θ ) s2 ⇔ (s2θ s1)Rs3 = s2θ s1Rs3 ⇔ (Rs3 ,I, Rs3 ): −1 ∗ ∗ −1 −1 ∗ (S, θ) → (S, (θ ) ) ⇔ (I, Rs3 , Rs3 ):(S, θ ) → (S, θ ) ⇔ (S, θ) is isotopic to (S, (θ ) ) ⇔ (S, θ∗) is isotopic to (S, θ−1). −1 −1 The proof of the equivalence of (3) and (4) is as follows. Rs2 Rs3 = Rs2θs3 ⇔ Rs2 Rs3 = R−1 ⇔ R R = R ⇔ (s −1θs )−1θs = s −1θ(s θs ) ⇔ (s θs )L = s L −1θs = s2θs3 s3 s2 s2θs3 1 3 2 1 2 3 2 3 s1 3 s1 2 −1 ∗ −1 ∗ ∗ −1 s2( θ) s3Ls1 ⇔ (s2θs3)Ls1 = s2( θ) s3Ls1 ⇔ (s3θ s2)Ls1 = s3Ls1 θs2 ⇔ (I, Ls1 , Ls1 ): −1 ∗ ∗ −1 −1 ∗ (S, θ) → (S, ( θ) ) ⇔ (Ls1 ,I, Ls1 ):(S, θ ) → (S, θ) ⇔ (S, θ) is isotopic to (S, ( θ) ) ⇔ (S, θ∗) is isotopic to (S, −1θ). Remark 3.1 In the proof of Theorem 3.1, it can be observed that the isotopisms are triples of the forms (A, I, A) and (I,B,B). All weak associative identities such as the Bol, Moufang and extra identities have been found to be isotopic invariant in loops for any triple of the form (A, B, C) while the central identities have been found to be isotopic invariant only under triples of the forms (A, B, A) and (A, B, B). Since associativity obeys all the Bol-Moufang identities, the observation in the theorem agrees with the latter stated facts. Corollary 3.1 (L, θ) is a Smarandache quasigroup with associative subquasigroup (S, θ) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Let (L, θ) be the quasigroup in consideration. By Lemma 2.2, (S, θ) is a group. Notice that R R = R ⇔ L∗ = L∗ L∗ . Hence, (S, θ∗) is also a group. In Theorem s2 s3 s2θs3 s2θs3 s3 s2 3.1, two of the parastrophes are isotopes of (S, θ) while the other two are isotopes of (S, θ∗). Vol. 2 Parastrophic invariance of Smarandache quasigroups 51

Since the Khalil conditions are neccessary and sufficient conditions for a quasigroup to be an isotope of a group, then they must be necessarily and sufficiently true in the four quasigroup parastrophes of (S, θ). Lemma 3.1 (L, θ∗) is a Smarandache quasigroup with associative subquasigroup (S, θ∗) if and only if any of the following equivalent statements is true. 1. (S, θ∗) is isotopic to (S, −1θ). 2. (S, θ) is isotopic to (S, (−1θ)∗). 3. (S, θ∗) is isotopic to (S, θ−1). 4. (S, θ) is isotopic to (S, (θ−1)∗). Proof. Replace (L, θ) with (L, θ∗) in Theorem 3.1. Corollary 3.2 (L, θ∗) is a Smarandache quasigroup with associative subquasigroup (S, θ∗) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Replace (L, θ) with (L, θ∗) in Corollary 3.1. Lemma 3.2 (L, θ−1) is a Smarandache quasigroup with associative subquasigroup (S, θ−1) if and only if any of the following equivalent statements is true. 1. (S, θ−1) is isotopic to (S, θ∗). 2. (S, (θ−1)∗) is isotopic to (S, θ). 3. (S, θ−1) is isotopic to (S, −1θ). 4. (S, (θ−1)∗) is isotopic to (S, (−1θ)∗). Proof. Replace (L, θ) with (L, θ−1) in Theorem 3.1. Corollary 3.3 (L, θ−1) is a Smarandache quasigroup with associative subquasigroup (S, θ−1) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Replace (L, θ) with (L, θ−1) in Corollary 3.1. Lemma 3.3 (L, −1θ) is a Smarandache quasigroup with associative subquasigroup (S, −1θ) if and only if any of the following equivalent statements is true. 1. (S, −1θ) is isotopic to (S, θ−1). 2. (S, (−1θ)∗) is isotopic to (S, (θ−1)∗). 3. (S, −1θ) is isotopic to (S, θ∗). 4. (S, (−1θ)∗) is isotopic to (S, θ). Proof. Replace (L, θ) with (L, −1θ) in Theorem 3.1. Corollary 3.4 (L, −1θ) is a Smarandache quasigroup with associative subquasigroup (S, −1θ) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Replace (L, θ) with (L, −1θ) in Corollary 3.1. Lemma 3.4 (L, (θ−1)∗) is a Smarandache quasigroup with associative subquasigroup (S, (θ−1)∗) if and only if any of the following equivalent statements is true. 1. (S, (θ−1)∗) is isotopic to (S, (−1θ)∗). 2. (S, θ−1) is isotopic to (S, −1θ). 3. (S, (θ−1)∗) is isotopic to (S, θ). 4. (S, θ−1)) is isotopic to (S, θ∗). Proof. Replace (L, θ) with (L, (θ−1)∗) in Theorem 3.1. 52 Tèm´ıtópéGbóláhànJa´ıyéo.lá No. 3

Corollary 3.5 (L, (θ−1)∗) is a Smarandache quasigroup with associative subquasigroup (S, (θ−1)∗) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Replace (L, θ) with (L, (θ−1)∗) in Corollary 3.1. Lemma 3.5 (L, (−1θ)∗) is a Smarandache quasigroup with associative subquasigroup (S, (−1θ)∗) if and only if any of the following equivalent statements is true. 1. (S, (−1θ)∗) is isotopic to (S, θ). 2. (S, −1θ) is isotopic to (S, θ∗). 3. (S, (−1θ)∗) is isotopic to (S, (θ−1)∗). 4. (S, −1θ) is isotopic to (S, θ−1). Proof. Replace (L, θ) with (L, (−1θ)∗) in Theorem 3.1. Corollary 3.6 (L, (−1θ)∗) is a Smarandache quasigroup with associative subquasigroup (S, (−1θ)∗) if and only if any of the six Khalil conditions is true for some four parastrophes of (S, θ). Proof. Replace (L, θ) with (L, (−1θ)∗) in Corollary 3.1.

Theorem 3.2 (L, πi) is a Smarandache quasigroup with associative subquasigroup (S, πi)

∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if for any of some four j ∈ {1, 2, 3, 4, 5, 6},(S, πj) is an isotope

of (S, πi) or (S, πk) for one k ∈ {1, 2, 3, 4, 5, 6} such that i 6= j 6= k. Proof. This is simply the summary of Theorem 3.1, Lemma 3.1, Lemma 3.2, Lemma 3.3, Lemma 3.4 and Lemma 3.5.

Corollary 3.7 (L, πi) is a Smarandache quasigroup with associative subquasigroup

(S, πi) ∀ i ∈ {1, 2, 3, 4, 5, 6} if and only if any of the six Khalil conditions is true for any of some four of (S, πi). Proof. This can be deduced from Theorem 3.2 and the Khalil conditions or by combining Corollary 3.1, Corollary 3.2, Corollary 3.3, Corollary 3.4, Corollary 3.5 and Corollary 3.6.

References

[1] R. Artzy, Isotopy and Parastrophy of Quasigroups, Proc. Amer. Math. Soc. 14(1963), pp. 429-431. [2] V. D. Belousov, Systems of quasigroups with generalised identities, Usp. Mat. Nauk., 20(1965), pp. 75-146. [3] R. H. Bruck, A survey of binary systems, Springer-Verlag, Berlin-G¨ottingen-Heidelberg, 966, pp. 185. [4] O. Chein, H. O. Pﬂugfelder and J. D. H. Smith, Quasigroups and Loops : Theory and Applications, Heldermann Verlag, 1990, pp. 568. [5] J. Dene and A. D. Keedwell, Latin squares and their applications, the English University press Lts, 1974, pp. 549. [6] R. A. Fisher and F. Yates, The 6×6 Latin squares, Proc. Camb. Philos. Soc., 30(1934), pp. 429-507. Vol. 2 Parastrophic invariance of Smarandache quasigroups 53

[7] E. G. Goodaire, E. Jespers and C. P. Milies, Alternative Loop Rings, NHMS(184), Elsevier, 1996, pp. 387. [8] K. Kunen, Quasigroups, Loops and Associative Laws, J. Alg., 185(1996), pp. 194-204. [9] A. S. Muktibodh, Smarandache quasigroups rings, Scientia Magna, 1(2005), No. 2, pp. 139-144. [10] A. S. Muktibodh, Smarandache quasigroups, Scientia Magna, 2(2006), No. 1, pp. 13-19. [11] H. O. Pﬂugfelder, Quasigroups and Loops : Introduction, Sigma series in Pure Math. 7, Heldermann Verlag, Berlin, 1990, pp. 147. [12] A. Sade, Quasigroupes parastrophiques, Math. Nachr., 20(1959), pp. 73-106. [13] K. Shahbazpour, On identities of isotopy closure of variety of groups, Milehigh confer- ence on loops, quasigroups and non-associative systems, University of Denver, Denver, Colorado, 2005. [14] S. K. Stein, Foundation of quasigroups, Proc. Nat. Acad. Sci., 42(1956), pp. 545-545. [15] S. K. Stein, On the foundation of quasigroups, Trans. Amer. Math. Soc., 85(1957), pp. 228-256. [16] W. B. Vasantha Kandasamy, Smarandache Loops, Department of Mathematics, Indian Institute of Technology, Madras, India, 2002, pp. 128. [17] W. B. Vasantha Kandasamy, Smarandache Loops, Smarandache notions journal, 13(2002), pp. 252-258. [18] W. B. Vasantha Kandasamy, Smarandache groupoids, American Research Press (to appear). Scientia Magna Vol. 2 (2006), No. 3, 54-64

Perfect powers in Smarandache n-Expressions

Muneer Jebreel Karama

SS-Math-Hebron (Hebron Education O–ce/UNRWA) Field Education O–cer, Jerusalem, Box 19149, Israel

Abstract In [1] I studied the concept of Smarandache n-expressions, for example I proposed formulas, found solutions, proposed open questions, and conjectured, but all for the fixed 3, and 2 numbers, but what will happen if these equations have different fixed numbers such as 7? This paper will answer this question. Keywords Perfect powers in Smarandache N-Expressions, perfect squares, identity.

§1. Introduction

In [4] M. Perez and E. Burton, documented that J. Castillo [5], asked how many primes are there in the Smarandache n-expressions:

x2 x3 x1 x1 + x2 + ··· + xn , where n > 1, x1, x2, ··· , xn > 1, and (x1, x2, ··· xn) = 1. In this paper, with only slight modiﬂcation of the above equation we got the following equation namely: ax1 + ax2 + ··· + axn , where a > 1, x1, x2, ··· , xn > 1, and (a, x1, x2, ··· , xn) = 1.

§2.Main results and proofs

I will study the following cases of above equation. Case 1. The solution of equation (1) is given by

7p + 7q + 7r + 7s = k2, (1) where p = 2m, q = 2m + 1, r = 2m + 2, s = 2m + 3, and k = 20 · 7m. Proof. Assume k = 20 · 7m, then k2 = 400 · 72m, i.e.

k2 = 400 · 72m = (1 + 7 + 49 + 343)72m = 72m + 72m+1 + 72m+2 + 72m+3.

Hence p = 2m, q = 2m + 1, r = 2m + 2, and s = 2m + 3. Vol. 2 Perfect powers in Smarandache n-Expressions 55

The ﬂrst 11th solution of (1) is given in Table 1 below: Table 1

m 7p + 7q + 7r + 7s k2 0 70 + 71 + 72 + 73 202 1 72 + 73 + 74 + 75 1402 2 74 + 75 + 76 + 77 9802 3 76 + 77 + 78 + 79 68602 4 78 + 79 + 710 + 711 480202 5 710 + 711 + 712 + 713 3361402 6 712 + 713 + 714 + 715 23529802 7 714 + 715 + 716 + 717 164708602 8 716 + 717 + 718 + 719 1152960202 9 718 + 719 + 720 + 721 8070721402 10 720 + 721 + 722 + 723 5649504982

The ﬂrst terms and the m-th terms of the sequence (the last column on Table 1 are:)

400, 9600, 960400, 4705900, ··· , 20 · 2 · 72m, ··· . (2)

20, 140, 980, 68600, ··· , 20 · 72m, ··· (3)

I have noticed there is no prime numbers in geometric series (3), (excluding the prime 7). The Xm 10(7m − 1) 20 · 72m = , 3 i 10(7m−1) and there is no limit, since 3 becomes large as m approach inﬂnity. The sequence has no limit, therefore it is divergent, but the summation of reciprocal convergent. Equation (1) has the following important properties, i.e.

7p + 7q + 7r + 7s = a2 − b2 = c2 − d2 = e2 − f 2 = g2 − h2 = i2 − k2 (4) = j2 − l2 = o2 − t2 = u2 − v2 = w2 − x2.

The solution of (4) are

a = 25 · 7n−1, b = 15 · 7n−1, c = 100 · 72n−2 + 1,

d = 100 · 72n−2 − 1, e = 50 · 72n−2 + 2, f = 50 · 72n−2 − 2,

g = 4 · 72n−2 + 25, h = 4 · 72n−2 − 25, i = 10 · 72n−1 + 10,

k = 10 · 72n−1 − 10, j = 50 · 72n−1 + 20, l = 50 · 72n−1 − 20,

o = 72n−1 + 100, t = 72n−1 − 100, u = 2 · 72n−1 + 50, 56 Muneer Jebreel Karama No. 3

v = 2 · 72n−1 − 50, w = 25 · 72n−1 + 4, x = 25 · 72n−1 − 4.

For example, let n = 2 (m = 1), then we always have n = m + 1.

1752 − 1052 = 49012 − 48992 = 24522 − 24482 = 2212 − 1712 = 5002 − 4802 = 2652 − 2252 = 1492 − 512 = 1482 − 482 = 12292 − 12212 = 1402 = 72 + 73 + 74 + 75.

Conjecture 1. If p, q, r, s are distinct prime numbers, then the equation

7p + 7q + 7r + 7s = k2, will has no solution, (otherwise we have solutions in prime numbers, such as

72 + 72 + 72 + 72 = 142, and 72 + 72 + 73 + 73 = 282.)

Case 2. The solution of equation (5) is given by

7p + 7q + 7r + 7s = k2, (5) where p = q = 2m, r = s = 2m + 1, and k = 4 · 7m. Proof. Assume k = 4 · 7m, then k2 = 16 · 72m, i.e.

k2 = 16 · 72m = (1 + 1 + 7 + 7)72m = 72m + 72m + 72m+1 + 72m+1.

Hence p = q = 2m, r = s = 2m + 1. The ﬂrst 11th solution of (5) is given in Table 2 below: Table 2

m 7p + 7q + 7r + 7s k2 0 70 + 70 + 71 + 71 42 1 72 + 72 + 73 + 73 282 2 74 + 74 + 75 + 75 1962 3 76 + 76 + 77 + 77 13722 4 78 + 78 + 79 + 79 96042 5 710 + 710 + 711 + 711 672282 6 712 + 712 + 713 + 713 4705962 7 714 + 714 + 715 + 715 32941722 8 716 + 716 + 717 + 717 230592042 9 718 + 718 + 719 + 719 1614144282 10 720 + 720 + 721 + 721 11299009962 Vol. 2 Perfect powers in Smarandache n-Expressions 57

Equation (5) has the following important properties, i.e.

7p + 7q + 7r + 7s = a2 − b2 = c2 − d2 = e2 − f 2 = g2 − h2, (6) where the solution of (6) are

a = 5 · 7n−1, b = 3 · 7n, c = 4 · 72n−2 + 1, d = 4 · 72n−2 − 1,

e = 2 · 72n−2 + 2, f = 2 · 72n−2 − 2, g = 72n−2 + 4, h = 72n−2 − 4. For example, let n = 2 (m = 1), then we have

352 − 212 = 1972 − 1952 = 1002 − 962 = 532 − 452 = 282 = 72 + 72 + 73 + 73.

Case 3. The solution of equation (7) is given by

7p + 7q + 7r + 7s = k2, (7) where p = q = 2m, r = s = 2m + 2, and k = 10 · 7m. Proof. Assume k = 10 · 7m, then k2 = 100 · 72m, i.e.

k2 = 100 · 72m = (1 + 1 + 49 + 49)72m = 72m + 72m + 72m+2 + 72m+2.

Hence p = q = 2m, r = s = 2m + 2. The ﬂrst 11th solution of (7) is given in Table 3 below: Table 3

m 7p + 7q + 7r + 7s k2 0 70 + 70 + 72 + 72 102 1 72 + 72 + 74 + 74 702 2 74 + 74 + 76 + 76 4902 3 76 + 76 + 78 + 78 34302 4 78 + 78 + 710 + 710 240102 5 710 + 710 + 712 + 712 1680702 6 712 + 712 + 714 + 714 11764902 7 714 + 714 + 716 + 716 82354302 8 716 + 716 + 718 + 718 576480102 9 718 + 718 + 720 + 720 4035360702 10 720 + 720 + 722 + 722 28247524902

Equation (7) has the following important properties, i.e.

7p + 7q + 7r + 7s = k2 = a2 + b2, (8) where the solution of (8) are a = 8 · 7n−1, b = 6 · 7n−1. 58 Muneer Jebreel Karama No. 3

For example, let n = 2 (m = 1), then we have

72 + 72 + 74 + 74 = 562 + 422 = 702.

Case 4. The solution of equation (9) is given by

7p + 7q + 7r + 7s = k2, (9) where p = q = r = s = 2m, and k = 2 · 7m. Proof. Assume k = 2 · 7m, then k2 = 4 · 72m, i.e.

k2 = 4 · 72m = (1 + 1 + 1 + 1)72m = 72m + 72m + 72m+2 + 72m+2.

Hence p = q = r = s = 2m. The ﬂrst 11th solution of (9) is given in Table 4 below: Table 4

m 7p + 7q + 7r + 7s k2 0 70 + 70 + 70 + 70 22 1 72 + 72 + 72 + 72 142 2 74 + 74 + 74 + 74 982 3 76 + 76 + 76 + 76 6862 4 78 + 78 + 78 + 78 48022 5 710 + 710 + 710 + 710 336142 6 712 + 712 + 712 + 712 2352982 7 714 + 714 + 714 + 714 16470682 8 716 + 716 + 716 + 716 115296022 9 718 + 718 + 718 + 718 807072142 10 720 + 720 + 720 + 720 5649504982

Equation (7) has the following important properties, i.e.

p q r s 2 2 2 2 7 + 7 + 7 + 7 = ki + ki+1 = a + a . (10)

m m where ki = 2 · 7 and a = 10 · 7 . Examples of equation (10): 1) 22 + 42 = 102 + 102, (divided both sides by 2 given 12 + 72 = 52 + 52) 2) 142 + 982 = 702 + 702, (divided both sides by 14 given 12 + 72 = 52 + 52) 3) 982 + 6862 = 4902 + 4902. (divided both sides by 98 given 12 + 72 = 52 + 52) These examples suggested a formula that gives three perfect squares which are in the arithmetic progression. So for the positive m and n with m > n, put

x = 2mn − m2 + n2,

y = m2 + n2, Vol. 2 Perfect powers in Smarandache n-Expressions 59

z = 2mn + m2 − n2, such as y2 + y2 = z2 + x2. So if m = 2, n = 1, then we will have

12 + 72 = 52 + 52.

Case 5. The solution of equation (11) is given by

7p + 7q + 7r + 7s = 24 · 52 · 7p, (11) where p = 2m + 1, q = 2m + 2, r = 2m + 3, and s = 2m + 4. Proof. Assume 24 · 52 · 72m+1, is the sum of equation (11), then

24 · 52 · 72m+1 = (1 + 7 + 49 + 343)72m+1 = 72m+1 + 72m+2 + 72m+3 + 72m+4.

Hence p = 2m + 1, q = 2m + 2, r = 2m + 3, s = 2m + 4. The ﬂrst 7th solution is given in Table 5 below: Table 5

m 7p + 7q + 7r + 7s 24 · 52 · 7p 0 71 + 72 + 73 + 74 24 · 52 · 71 1 73 + 74 + 75 + 76 24 · 52 · 73 2 75 + 76 + 77 + 78 24 · 52 · 75 3 77 + 78 + 79 + 710 24 · 52 · 77 4 79 + 710 + 711 + 712 24 · 52 · 79 5 711 + 712 + 713 + 714 24 · 52 · 711 6 713 + 714 + 715 + 716 24 · 52 · 713

The ﬂrst terms and the m-th terms of the sequence (the last column on Table 5 are:)

24 · 52 · 71, 24 · 52 · 73, 24 · 52 · 75, ··· , 24 · 52 · 72m+1, ··· , (12) where the square roots are

20 · 71/2, 20 · 73/2, 20 · 75/2, 20 · 77/2, ··· , 20 · 72m+1/2, ··· . (13)

Notice that there is no prime numbers in geometric series (13). The Xm 10 · 71/2 · 72m+1 − 1 20 · 72i+1/2 = , 3 i 10·71/2·72m+1−1 and there is no limit, since 3 becomes large as m approach inﬂnity. The sequence has no limit, therefore it is divergent, but the summation of reciprocal convergent. Conjecture 2. If p, q, r, s are distinct prime numbers, then the equation

7p + 7q + 7r + 7s = 24 · 52 · 7p, 60 Muneer Jebreel Karama No. 3 has no solution. Case 6. The solution of equation (11) is given by

7p + 7q + 7r + 7s = 24 · 7p, (14) where p = q = 2m + 1, r = s = 2m + 2. Proof. Assume 24 · 72m+1 is the sum of equation (14), then

24 · 72m+1 = (1 + 1 + 7 + 7)72m+1 = 72m+1 + 72m+1 + 72m+2 + 72m+2.

Hence p = q = 2m + 1, r = s = 2m + 2. The ﬂrst 11th solution is given in Table 6 below: Table 6

m 7p + 7q + 7r + 7s 24 · 7p 0 71 + 71 + 72 + 72 24 · 71 1 73 + 73 + 74 + 74 24 · 73 2 75 + 75 + 76 + 76 24 · 75 3 77 + 77 + 78 + 78 24 · 77 4 79 + 79 + 710 + 710 24 · 79 5 711 + 711 + 712 + 712 24 · 711 6 713 + 713 + 714 + 714 24 · 713 7 715 + 715 + 716 + 716 24 · 715 8 717 + 717 + 718 + 718 24 · 717 9 719 + 719 + 720 + 720 24 · 719 10 721 + 721 + 722 + 722 24 · 721

If we look deeply in equation (11), (14), we can ﬂnd the following relation

204 − 24 = 972 − 952 = 502 − 462 = 282 − 202.

Case 7. The solution of equation (15) is given by

7p + 7q + 7r + 7s = 22 · 52 · 7p, (15) where p = q = 2m + 1, r = s = 2m + 3. Proof. Assume 22 · 52 · 72m+1, is the sum of equation (15), then

22 · 52 · 72m+1 = (1 + 1 + 49 + 49)72m+1 = 72m+1 + 72m+1 + 72m+3 + 72m+3.

Hence p = q = 2m + 1, r = s = 2m + 2. The ﬂrst 11th solution of (15) is given in Table 7 below: Table 7 Vol. 2 Perfect powers in Smarandache n-Expressions 61

m 7p + 7q + 7r + 7s 22 · 52 · 7p 0 71 + 71 + 73 + 73 22 · 52 · 71 1 73 + 73 + 75 + 75 22 · 52 · 73 2 75 + 75 + 77 + 77 22 · 52 · 75 3 77 + 77 + 79 + 79 22 · 52 · 77 4 79 + 79 + 711 + 711 22 · 52 · 79 5 711 + 711 + 713 + 713 22 · 52 · 711 6 713 + 713 + 715 + 715 22 · 52 · 713 7 715 + 715 + 717 + 717 22 · 52 · 715 8 717 + 717 + 719 + 719 22 · 52 · 717 9 719 + 719 + 721 + 721 22 · 52 · 719 10 721 + 721 + 723 + 723 22 · 52 · 721

case 8. The solution of equation (16) is given by

7p + 7q + 7r + 7s = 1201 · 22 · 52 · 7p, (16) where p = 2m, q = 2m + 2, r = 2m + 4, s = 2m + 6. Proof. Assume 1201 · 22 · 52 · 72m+1, is the sum of equation (16), then

1201 · 22 · 52 · 72m+1 = (1 + 49 + 2401 + 117649)72m = 72m + 72m+2 + 72m+4 + 72m+6.

Hence p = 2m, q = 2m + 2, r = 2m + 4, s = 2m + 6. The ﬂrst 11th solution of (16) is given in Table 8 below: Table 8

m 7p + 7q + 7r + 7s 1201 · 22 · 52 · 7p 0 70 + 72 + 74 + 76 1201 · 22 · 52 · 70 1 72 + 74 + 76 + 78 1201 · 22 · 52 · 72 2 74 + 76 + 78 + 710 1201 · 22 · 52 · 74 3 76 + 78 + 710 + 712 1201 · 22 · 52 · 76 4 78 + 710 + 712 + 714 1201 · 22 · 52 · 78 5 710 + 712 + 714 + 716 1201 · 22 · 52 · 710 6 712 + 714 + 716 + 718 1201 · 22 · 52 · 712 7 714 + 716 + 718 + 720 1201 · 22 · 52 · 714 8 716 + 718 + 720 + 722 1201 · 22 · 52 · 716 9 718 + 720 + 722 + 724 1201 · 22 · 52 · 718 10 720 + 722 + 724 + 726 1201 · 22 · 52 · 720

Case 9. The solution of equation (17) is given by

7p + 7q + 7r + 7s = 1201 · 22 · 52 · 7p, (17) 62 Muneer Jebreel Karama No. 3 where p = 2m + 1, q = 2m + 3, r = 2m + 5, s = 2m + 7. Proof. Assume 1201 · 22 · 52 · 72m+1, is the sum of equation (17), then

1201 · 22 · 52 · 72m+1 = (1 + 49 + 2401 + 117649)72m+1 = 72m+1 + 72m+3 + 72m+5 + 72m+7.

Hence p = 2m + 1, q = 2m + 3, r = 2m + 5, s = 2m + 7. The ﬂrst 11th solution of (17) is given in Table 9 below: Table 9

m 7p + 7q + 7r + 7s 1201 · 22 · 52 · 7p 0 71 + 73 + 75 + 77 1201 · 22 · 52 · 71 1 73 + 75 + 77 + 79 1201 · 22 · 52 · 73 2 75 + 77 + 79 + 711 1201 · 22 · 52 · 75 3 77 + 79 + 711 + 713 1201 · 22 · 52 · 77 4 79 + 711 + 713 + 715 1201 · 22 · 52 · 79 5 711 + 713 + 715 + 717 1201 · 22 · 52 · 711 6 713 + 715 + 717 + 719 1201 · 22 · 52 · 713 7 715 + 717 + 719 + 721 1201 · 22 · 52 · 715 8 717 + 719 + 721 + 723 1201 · 22 · 52 · 717 9 719 + 721 + 723 + 725 1201 · 22 · 52 · 719 10 721 + 723 + 725 + 727 1201 · 22 · 52 · 721

Case 10. The solution of equation (18) is given by

7p + 7q + 7r + 7s + 7t + 7u + 7v + 7w + 7z = 1201 · 25 · 52 · 7p, (18) where p = m, q = m + 1, r = m + 2, s = m + 3, t = m + 4, u = m + 5, w = m + 6, z = m + 7. Proof. Assume 1201 · 22 · 52 · 72m+1, is the sum of equation (18), then

1201 · 22 · 52 · 72m+1 = (1 + 49 + 2401 + 117649)72m+1 = 72m+1 + 72m+3 + 72m+5 + 72m+7.

Hence p = m, q = m + 1, r = m + 2, s = m + 3, t = m + 4, u = m + 5, w = m + 6, z = m + 7. The ﬂrst 11th solution of (18) is given in Table 10 below: Table 10 Vol. 2 Perfect powers in Smarandache n-Expressions 63

m 7p + 7q + 7r + 7s + 7t + 7u + 7w + 7z 1201 · 25 · 52 · 7p 0 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 1201 · 25 · 52 · 70 1 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 1201 · 25 · 52 · 71 2 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79 1201 · 25 · 52 · 72 3 73 + 74 + 75 + 76 + 77 + 78 + 79 + 710 1201 · 25 · 52 · 73 4 74 + 75 + 76 + 77 + 78 + 79 + 710 + 711 1201 · 25 · 52 · 74 5 75 + 76 + 77 + 78 + 79 + 710 + 711 + 712 1201 · 25 · 52 · 75 6 76 + 77 + 78 + 79 + 710 + 711 + 712 + 713 1201 · 25 · 52 · 76 7 77 + 78 + 79 + 710 + 711 + 712 + 713 + 714 1201 · 25 · 52 · 77 8 78 + 79 + 710 + 711 + 712 + 713 + 714 + 715 1201 · 25 · 52 · 78 9 79 + 710 + 711 + 712 + 713 + 714 + 715 + 716 1201 · 25 · 52 · 79 10 710 + 711 + 712 + 713 + 714 + 715 + 716 + 717 1201 · 25 · 52 · 710

Case 11. The solution of equation (19) is given by

7p + 7q + 7r + 7s + 7t + 7u + 7v + 7w + 7z = k3, (19) where p = q = r = s = t = u = w = z = 3m. Proof. Assume k = 2 · 7m, then k3 = 23 · 73m, i.e.

k2 = 8 · 73m = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)73m = 73m + 73m + 73m + 73m + 73m + 73m + 73m + 73m.

Hence p = q = r = s = t = u = w = z = 3m. The ﬂrst 11th solution of (19) is given in Table 11 below: Table 11

m 7p + 7q + 7r + 7s + 7t + 7u + 7w + 7z k3 = 23 · 73m 0 70 + 70 + 70 + 70 + 70 + 70 + 70 + 70 23 1 73 + 73 + 73 + 73 + 73 + 73 + 73 + 73 143 2 76 + 76 + 76 + 76 + 76 + 76 + 76 + 76 983 3 79 + 79 + 79 + 79 + 79 + 79 + 79 + 79 6863 4 712 + 712 + 712 + 712 + 712 + 712 + 712 + 712 48023 5 715 + 715 + 715 + 715 + 715 + 715 + 715 + 715 336143 6 718 + 718 + 718 + 718 + 718 + 718 + 718 + 718 2352983 7 721 + 721 + 721 + 721 + 721 + 721 + 721 + 721 16470683 8 724 + 724 + 724 + 724 + 724 + 724 + 724 + 724 115296023 9 727 + 727 + 727 + 727 + 727 + 727 + 727 + 727 807072143 10 730 + 730 + 730 + 730 + 730 + 730 + 730 + 730 5649504983

I ﬂnd interesting identity like this: 64 Muneer Jebreel Karama No. 3

1) 143 − 23 = 6852 − 6832 = 3442 − 3402 = 1752 − 1672, 2) 983 − 143 = 2346132 − 2346112 = 1173082 − 1173042 = 586572 − 586492, and so on.

§3. Open questions

The following equations have inflnitely many solutions, flnd them. 1) −7p + 7q − 7r + 7s = 3 · 22 · 52 · 7p ( has two difierent solutions), 2) 7p + 7q = 23 · 7p, 3) 7p + 7q = 2 · 52 · 7p.

References

[1] Karama. J. Muneer, Perfect Powers in Smarandache n-Expressions, Scientia Magna, 1(2005), pp. 15-24. [2] http://www.gallup.unm.edu/ smarandache. [3] E. Burton, M. Perez, Some Notions and Questions in Number Theory, Department of Mathematics 3400 Cluj-Napoca, Babes- Bolyai University, Romania, SNJ, 3(1993). [4] J. Castillo, The Smarandache n-Expressions, Mathematical Spectrum, 29(1997/8), pp. 21. Scientia Magna Vol. 2 (2006), No. 3, 65-77

Palindromic permutations and generalized Smarandache palindromic permutations

Tèm´ıtópé Gbóláhán Ja´ıyéo.lá 1 Department of Mathematics, Obafemi Awolowo University, Ile Ife, Nigeria.

Abstract The idea of left(right) palindromic permutations(LPPs, RPPs) and left(right) generalized Smarandache palindromic permutations(LGSPPs, RGSPPs) are introduced in symmetric groups Sn of degree n. It is shown that in Sn, there exist a LPP and a RPP and they are unique(this fact is demonstrated using S2 and S3). The dihedral group Dn is shown to be generated by a RGSPP and a LGSPP(this is observed to be true in S3) but the geometric interpretations of a RGSPP and a LGSPP are found not to be rotation and reﬂection respectively. In S3, each permutation is at least a RGSPP or a LGSPP. There are 4 RGSPPs and 4 LGSPPs in S3, while 2 permutations are both RGSPPs and LGSPPs. A permutation in Sn is shown to be a LPP or RPP(LGSPP or RGSPP) if and only if its inverse is a LPP or RPP(LGSPP or RGSPP) respectively. Problems for future studies are raised. Keywords Permutation, symmetric groups, palindromic permutations, generalized Smarandache palindromic permutations.

§1. Introduction

According to Ashbacher and Neirynck [1], an integer is said to be a palindrome if it reads the same forwards and backwards. For example, 12321 is a palindromic number. They also stated that it is easy to prove that the density of the palindromes is zero in the set of positive integers and they went ahead to answer the question on the density of generalized Smarandache palindromes (GSPs) by showing that the density of GSPs in the positive integers is approxi- mately 0.11. Gregory [2], Smarandache [8] and Ramsharan [7] deﬁned generalized Smarandache palindrome (GSP) as any integer or number of the form

a1a2a3 · · · anan · · · a3a2a1 or a1a2a3 · · · an−1anan−1 · · · a3a2a1

where all a1, a2, a3, · · · , an ∈ N having one or more digits. On the other hand, Hu [3] calls any integer or number of this form a Smarandache generalized palindrome (SGP). His naming will not be used here the ﬁrst naming will be adopted. Numbers of this form have also been considered by Khoshnevisan [4], [5] and [6]. For the sake of clari cation, it must be mentioned that the possibility of the trivial case of enclosing the entire number is excluded. For example, 12345 can be written as (12345). In this case, the number is simply said to be a palindrome or a palindromic number as it was mentioned earlier on. So, every number is a GSP. But this possibility is eliminated by requiring that each number be split into at least two segments if it is not a regular palindrome. Trivially, since each regular palindrome is also

1On Doctorate Programme at the University of Agriculture Abeokuta, Nigeria. 66 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

a GSP and there are GSPs that are not regular palindromes, there are more GSPs than there are regular palindromes. As mentioned by Gregory [2], very interesting GSPs are formed from smarandacheian sequences. For an illustration he cited the smarandacheian sequence

11, 1221, 123321, . . ., 123456789987654321, 1234567891010987654321,

12345678910111110987654321, . . . and observed that all terms are all GSPs. He also mentioned that it has been proved that the GSP 1234567891010987654321 is a prime and concluded his work by possing the question of How many primes are in the GSP sequence above? Special mappings such as morphisms(homomorphisms, endomorphisms, automorphisms, isomorphisms e.t.c) have been useful in the study of the properties of most algebraic structures(e.g groupoids, quasigroups, loops, semigroups, groups e.tc.). In this work, the notion of palindromic permutations and generalized Smarandache palindromic permutations are introduced and studied using the symmetric group on the set N and this can now be viewed as the study of some palindromes and generalized Smarandache palindromes of numbers. The idea of left(right) palindromic permutations(LPPs, RPPs) and left(right) generalized Smarandache palindromic permutations(LGSPPs, RGSPPs) are introduced in symmetric

groups Sn of degree n. It is shown that in Sn, there exist a LPP and a RPP and they are

unique. The dihedral group Dn is shown to be generated by a RGSPP and a LGSPP but the geometric interpretations of a RGSPP and a LGSPP are found not to be rotation and reﬂection

respectively. In S3, each permutation is at least a RGSPP or a LGSPP. There are 4 RGSPPs

and 4 LGSPPs in S3, while 2 permutations are both RGSPPs and LGSPPs. A permutation

in Sn is shown to be a LPP or RPP(LGSPP or RGSPP) if and only if its inverse is a LPP or

RPP(LGSPP or RGSPP) respectively. Some of these results are demonstrated with S2 and S3. Problems for future studies are raised. But before then, some deﬁnitions and basic results on symmetric groups in classical group theory which shall be employed and used are highlighted ﬁrst.

§2. Preliminaries

Deﬁnition 2.1 Let X be a non-empty set. The group of all permutations of X under

composition of mappings is called the symmetric group on X and is denoted by SX . A subgroup

of SX is called a permutation group on X.

It is easily seen that a bijection X ' Y induces in a natural way an isomorphism SX =∼ SY .

If |X| = n, SX is denoted by Sn and called the symmetric group of degree n.

A permutation σ ∈ Sn can be exhibited in the form

 1 2 · · · n   σ(1) σ(2) · · · σ(n)  consisting of two rows of integers; the top row has integers 1, 2, . . . , n, usually (but not necessarily) in their natural order, and the bottom row has σ(i) below i for each i = 1, 2, . . . , n. Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 67

This is called a two-row notation for a permutation. There is a simpler, one-row notation for a special kind of permutation called cycle.

Deﬁnition 2.2 Let σ ∈ Sn. If there exists a list of distinct integers x1, . . . , xr ∈ N such that

σ(xi) = xi+1, i = 1, . . . , r − 1,

σ(xr) = x1,

σ(x) = x if x ∈/ {x1, · · · , xr},

then σ is called a cycle of length r and denoted by (x1 . . . xr).

Remark 2.1 A cycle of length 2 is called a transposition. In other words , a cycle (x1 . . . xr)

moves the integers (x1 . . . xr) one step around a circle and leaves every other integer in N. If σ(x) = x, we say σ does not move x. Trivially, any cycle of length 1 is the identity mapping I or e. Note that the one-row notation for a cycle does not indicate the degree n, which has to be understood from the context. Deﬁnition 2.3 Let X be a set of points in space, so that the distance d(x, y) between points x and y is given for all x, y ∈ X. A permutation σ of X is called a symmetry of X if

d(σ(x), σ(y)) = d(x, y) ∀ x, y ∈ X.

Let X be the set of points on the vertices of a regular polygon which are labelled 1, 2, · · · , n i.e X = {1, 2, · · · , n}.

The group of symmetries of a regular polygon Pn of n sides is called the dihedral group of

degree n and denoted Dn.

Remark 2.2 It must be noted that Dn is a subgroup of Sn i.e Dn ≤ Sn.

Deﬁnition 2.4 Let Sn be a symmetric group of degree n. If σ ∈ Sn such that

1 2 · · · n σ =    σ(1) σ(2) · · · σ(n) 

then

1. the number Nλ(σ) = 12 . . . nσ(n) . . . σ(1) is called the left palindromic value(LP V ) of σ.

2. the number Nρ(σ) = 12 . . . nσ(1) . . . σ(n) is called the right palindromic value(RP V ) of σ.

Deﬁnition 2.5 Let σ ∈ SX such that

x1 x2 · · · xn σ =    σ(x1) σ(x2) · · · σ(xn) 

If X = N, then

1. σ is called a left palindromic permutation(LP P ) if and only if the number Nλ(σ) is a palindrome.

P Pλ(SX ) = {σ ∈ SX : σ is a LPP } 68 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

2. σ is called a right palindromic permutation(RP P ) if and only if the number Nρ(σ) is a palindrome.

P Pρ(SX ) = {σ ∈ SX : σ is a RPP }

3. σ is called a palindromic permutation(P P ) if and only if it is both a LP P and a RP P .

P P (SX ) = {σ ∈ SX : σ is a LPP and a RP P } = P Pλ(SX ) ∩ P Pρ(SX )

Deﬁnition 2.6 Let σ ∈ SX such that

x1 x2 · · · xn σ =    σ(x1) σ(x2) · · · σ(xn) 

If X = N, then 1. σ is called a left generalized Smarandache palindromic permutation(LGSP P ) if and only if the number Nλ is a GSP .

GSP Pλ(SX ) = {σ ∈ SX : σ is a LGSP P }

2. σ is called a right generalized Smarandache palindromic permutation(RGSP P ) if and only if the number Nρ is a GSP.

GSP Pρ(SX ) = {σ ∈ SX : σ is a RGSPP}

3. σ is called a generalized Smarandache palindromic permutation(GSP P ) if and only if it is both a LGSP P and a RGSP P .

GSP P (SX ) = {σ ∈ SX : σ is a LGSPP and a RGSP P } = GSP Pλ(SX ) ∩ GSP Pρ(SX )

Theorem 2.1 (Cayley Theorem) Every group is isomorphic to a permutation group.

Theorem 2.2 The dihedral group Dn is a group of order 2n generated by two elements σ, τ satisfying σn = e = τ 2 and τσ = σn−1τ, where

σ = 1 2 · · · n , and 1 2 · · · n τ =   .  1 n · · · 2 

§3. Main Results

Theorem 3.1 In any symmetric group Sn of degree n, there exists 1. a LP P and it is unique. 2. a RP P and it is unique. But there does not exist a P P .

Proof Let σ ∈ Sn, then Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 69

x1 x2 · · · xn σ =   .  σ(x1) σ(x2) · · · σ(xn)  1. When σ(n) = n, σ(n − 1) = n − 1, · · · , σ(2) = 2, σ(1) = 1 then the number

Nλ(σ) = 12 · · · nσ(n) · · · σ(2)σ(1) = 12 · · · nn · · · 21 is a palindrome which implies σ ∈ P Pλ(Sn). So there exist a LPP. The uniqueness is as follows. Observe that 1 2 · · · n σ =   = I.  1 2 · · · n 

Since Sn is a group for all n ∈ N and I is the identity element (mapping), then it must be unique. 2. When σ(1) = n, σ(2) = n − 1, · · · , σ(n − 1) = 2, σ(n) = 1, then the number

Nρ(σ) = 12 · · · nσ(1) · · · σ(n − 1)σ(n) = 12 · · · nn · · · 21 is a palindrome which implies σ ∈ P PρSn. So there exist a RPP. The uniqueness is as follows.

If there exist two of such, say σ1 and σ2 in Sn, then

 1 2 · · · n   1 2 · · · n  σ1 = and σ2 =  σ1(1) σ1(2) · · · σ1(n)   σ1(1) σ1(2) · · · σ1(n)  such that

Nρ(σ1) = 12 · · · nσ1(1) · · · σ1(n − 1)σ1(n) and

Nρ(σ2) = 12 · · · nσ2(1) · · · σ2(n − 1)σ2(n) are palindromes which implies

σ1(1) = n, σ1(2) = n − 1, · · · , σ1(n − 1) = 2, σ1(n) = 1, and

σ2(1) = n, σ2(2) = n − 1, · · · , σ2(n − 1) = 2, σ2(n) = 1.

So, σ1 = σ2, thus σ is unique. The proof of the last part is as follows. Let us assume by contradiction that there exists a

PP σ ∈ Sn. Then if 70 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

1 2 · · · n σ =   ,  σ(1) σ(2) · · · σ(n) 

Nλ(σ) = 12 · · · nσ(n) · · · σ(2)σ(1)

and

Nρ(σ) = 12 · · · nσ(1) · · · σ(n − 1)σ(n)

are palindromes. So that σ ∈ Sn is a PP. Consequently,

n = σ(n) = 1, n − 1 = σ(n − 1) = 2, · · · , 1 = σ(1) = n,

so that σ is not a bijection which means σ ∈/ Sn. This is a contradiction. Hence, no PP exist.

Example 3.1 Let us consider the symmetric group S2 of degree 2. There are two permutations of the set {1, 2} given by

1 2 I =   ,  1 2  and 1 2 δ =   .  2 1 

Nρ(I) = 1212 = (12)(12), Nλ(I) = 1221 or Nλ(I) = 1(22)1,

Nρ(δ) = 1221 or Nρ(δ) = (12)(21) and Nλ(δ) = 1212 = (12)(12).

So, I and δ are both RGSPPs and LGSPPs which implies I and δ are GSPPs i.e I, δ ∈

GSP Pρ(S2) and I, δ ∈ GSP Pρ(S2) ⇒ I, δ ∈ GSP P (S2). Therefore, GSP P (S2) = S2. Fur- thermore, it can be seen that the result in Theorem 3.1 is true for S2 because only I is a LPP and only δ is a RPP. There is no PP as the theorem says.

Example 3.2 Let us consider the symmetric group S3 of degree 3. There are six permutations of the set {1,2,3} given by

1 2 3 e = I =   ,  1 2 3 

 1 2 3  σ1 = ,  2 3 1 

 1 2 3  σ2 = ,  3 1 2  Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 71

1 2 3 e = I =   ,  1 2 3 

 1 2 3  τ1 = ,  1 3 2 

 1 2 3  τ2 = ,  3 2 1  and

 1 2 3  τ3 = .  2 1 3 

As claimed in Theorem 3.1, the unique LPP in S3 is I while the unique RPP in S3 is τ2. There is no PP as the theorem says.

Lemma 3.1 In S3, the following are true.

1. At least σ ∈ GSP Pρ(S3) or σ ∈ GSP Pλ(S3) ∀ σ ∈ S3.

2. |GSP Pρ(S3)| = 4, |GSP Pλ(S3)| = 4 and |GSP P (S3)| = 2. Proof Observe the following:

Nλ(I) = 123321, Nρ(I) = 123123 = (123)(123).

Nλ(σ1) = 123132, Nρ(σ1) = 123231 = 1(23)(23)1.

Nλ(σ2) = 123213, Nρ(σ2) = 123312 = (12)(33)(12).

Nλ(τ1) = 123231 = 1(23)(23)1, Nρ(τ1) = 123132.

Nλ(τ2) = 123123 = (123)(123), Nρ(τ2) = 123321.

Nλ(τ3) = 123312 = (12)(33)(12), Nρ(τ3) = 123213.

So, GSP Pλ(S3) = {I, τ1, τ2, τ3} and GSP Pρ(S3) = {I, σ1, σ2, τ2}. Thus, 1. is true .There-

Lemma 3.2 S3 is generated by a RGSPP and a LGSPP. Proof Recall from Example 3.2 that

S3 = {I, e, σ1, σ2, τ1, τ2, τ3}.

If σ = σ1 and τ = τ1, then it is easy to verify that

2 3 2 2 σ = σ2, σ = e, τ = e, στ = τ3, σ τ = τ2 = τσ hence,

2 2 S3 = {e, σ, σ , τ, στ, σ τ3} ⇒ S3 = hσ, τi. 72 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

From the proof Lemma 3.1, σ is a RGSPP and τ is a LGSPP. This justiﬁes the claim.

Remark 3.1 In Lemma 3.2, S3 is generated by a RGSPP and a LGSPP. Could this

statement be true for all Sn of degree n? Or could it be true for some subgroups of Sn? Also, it is interesting to know the geometric meaning of a RGSPP and a LGSPP. So two questions are possed and the two are answered.

Question 3.1 1. Is the symmetric group Sn of degree n generated by a RGSPP and a LGSPP? If not, what permutation group (s) is generated by a RGSPP and a LGSPP? 2. Are the geometric interpretations of a RGSPP and a LGSPP rotation and reﬂection respectively?

Theorem 3.2 The dihedral group Dn is generated by a RGSPP and a LGSPP i.e Dn =

hσ, τi where σ ∈ GSP Pρ(Sn) and τ ∈ GSP Pλ(Sn).

Proof Recall from Theorem 2.2 that the dihedral group Dn = hσ, τi where

 1 2 · · · n  σ = 1 2 · · · n =  2 3 · · · 1  and 1 2 · · · n τ =   .  1 n · · · 2  Observe that

Nρ(σ) = 123 · · · n23 · · · n1 = 1(23 · · · n)(23 · · · n)1, Nλ(σ) = 123 · · · n1n · · · 32.

Nρ(τ) = 12 · · · n1n · · · 2, Nλ(τ) = 12 · · · n2 · · · n1 = 1(2 · · · n)(2 · · · n)1.

So, σ ∈ GSP Pρ(Sn) and τ ∈ GSP Pλ(Sn). Therefore, the dihedral group Dn is generated by a RGSPP and a LGSPP.

Remark 3.2 In Lemma 3.2, it was shown that S3 is generated by a RGSPP and a LGSPP.

Considering Theorem 3.2 when n = 3, it can be deduced that D3 will be generated by a RGSPP

and a LGSPP. Recall that |D3| = 2 × 3 = 6, so S3 = D3. Thus Theorem 3.2 generalizes Lemma 3.2. Rotations and Reflections Geometrically, in Theorem 3.2, σ is a rotation of the regular 2π polygon Pn through an angle n in its own plane, and τ is a reflection (or a turning over) in the diameter through the vertex 1. It looks like a RGSPP and a LGSPP are formed by rotation and reflection respectively. But there is a contradiction in S4 which can be traced from a subgroup of S4 particularly the Klein four-group. The Klein four-group is the group of symmetries of a four sided non-regular polygon(rectangle). The elements are:

1 2 3 4 e = I =   ,  1 2 3 4 

 1 2 3 4  δ1 = ,  3 4 1 2  Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 73

 1 2 3 4  δ2 = ,  2 1 4 3  and  1 2 3 4  δ3 = .  4 3 2 1  Observe the following:

Nρ(δ1) = 12343412 = (12)(34)(34)(12), Nλ(δ1) = 12342143.

Nρ(δ2) = 12342143, Nλ(δ2) = 12343412 = (12)(34)(34)(12).

Nρ(δ3) = 12344321 = 123(44)321, Nλ(δ3) = 12341234 = (1234)(1234).

So, δ1 is a RGSPP while δ2 is a LGSPP and δ3 is a GSPP. Geometrically, δ1 is a rotation through

an angle of π while δ2 and δ3 are reﬂections in the axes of symmetry parallel to the sides. Thus

δ3 which is a GSPP is both a reflection and a rotation, which is impossible. Therefore, the geometric meaning of a RGSPP and a LGSPP are not rotation and reflection respectively. It is difficult to really ascertain the geometric meaning of a RGSPP and a LGSPP if at all it exist.

How beautiful will it be if GSP Pρ(Sn), P Pρ(Sn), GSP Pλ(Sn), P Pλ(Sn), GSP P (Sn) and

P P (Sn) form algebraic structures under the operation of map composition.

Theorem 3.3 Let Sn be a symmetric group of degree n. If σ ∈ Sn, then −1 1. σ ∈ P Pλ(Sn) ⇔ σ ∈ P Pλ(Sn). −1 2. σ ∈ P Pρ(Sn) ⇔ σ ∈ P Pρ(Sn).

3. I ∈ P Pλ(Sn).

Proof 1. σ ∈ P Pλ(Sn) implies

Nλ(σ) = 12 · · · nσ(n) · · · σ(2)σ(1)

is a palindrome. Consequently,

σ(n) = n, σ(n − 1) = n − 1, · · · , σ(2) = 2, σ(1) = 1.

−1 −1 Nλ(σ ) = σ(1)σ(2) · · · σ(n)n · · · 21 = 12 · · · nn · · · 21 ⇒ σ ∈ P Pλ(Sn).

The converse is similarly proved by carrying out the reverse of the procedure above.

2. σ ∈ P Pρ(Sn) implies

Nρ(σ) = 12 · · · nσ(1) · · · σ(n − 1)σ(n)

is palindrome. Consequently,

σ(1) = n, σ(2) = n − 1, · · · , σ(n − 1) = 2, σ(n) = 1. 74 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

−1 −1 Nρ(σ ) = σ(1)σ(2) · · · σ(n − 1)σ(n)12 · · · n = n · · · 2112 · · · n ⇒ σ ∈ P Pρ(Sn).

The converse is similarly proved by carrying out the reverse of the procedure above. 3. 1 2 · · · n I =   .  1 2 · · · n 

Nλ(I) = 12 · · · nn · · · 21 ⇒ I ∈ P Pλ(Sn).

Theorem 3.4 Let Sn be a symmetric group of degree n. If σ ∈ Sn, then −1 1.σ ∈ GSP Pλ(Sn) ⇐⇒ σ ∈ GSP Pλ(Sn). −1 2.σ ∈ GSP Pρ(Sn) ⇐⇒ σ ∈ GSP Pρ(Sn).

3.I ∈ GSP P (Sn).

Proof If σ ∈ Sn, then

1 2 · · · n σ =   .  σ(1) σ(2) · · · σ(n)  So

Nλ(σ) = 12 · · · nσ(n) · · · σ(2)σ(1) and

Nρ(σ) = 12 · · · nσ(1) · · · σ(n − 1)σ(n) are numbers with even number of digits whether n is an even or odd number. Thus, Nρ(σ) and

Nλ(σ) are GSPs deﬁned by

a1a2a3 · · · anan · · · a3a2a1 and not

a1a2a3 · · · an−1anan−1 · · · a3a2a1 where all a1, a2, a3, · · · , an ∈ N having one or more digits because the ﬁrst has even number of digits (or grouped digits) while the second has odd number of digits (or grouped digits). The following grouping notations will be used:

n (ai)i=1 = a1a2a3 · · · an and n [ai]i=1 = anan−1an−2 · · · a3a2a1.

Let σ ∈ Sn such that

x1 x2 · · · xn σ =   .  σ(x1) σ(x2) · · · σ(xn)  where xi ∈ N, ∀ i ∈ N.

1. So σ ∈ GSP Pλ(Sn) implies

n1 n2 n3 nn−1 nn Nλ(σ) = (xi1 ) (xi2 ) (xi3 ) · · · (xi −1 ) (xi ) i1=1 i2=(n1+1) i3=(n2+1) n in−1=(nn−2+1) n in=(nn−1+1) Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 75

nn nn−1 n3 n2 n1 [σ(x )] [σ(x −1 )] · · · [σ(x 3 )] [σ(x 2 )] [σ(x 1 )] in in=(nn−1+1) in in−1=(nn−2+1) i i3=(n2+1) i i2=(n1+1) i i1=1 N N N is a GSP, where xij ∈ , ∀ ij ∈ , j ∈ and nn = n. The interval of integers [1, n] is partitioned into

[1, n] = [1, n1] ∪ [n1 + 1, n2] ∪ · · · ∪ [nn−2 + 1, nn−1] ∪ [nn−1, nn].

nj nj The length of each grouping (·)ij or [·]ij is determined by the corresponding interval of integers [ni + 1, ni+1] and it is a matter of choice in other to make the number Nλ(σ) a GSP.

Now that Nλ(σ) is a GSP, the following are true:

(x )nn = [σ(x )]nn ⇔ [x ]nn = (σ(x ))nn in in=(nn−1+1) in in=(nn−1+1) in in=(nn−1+1) in in=(nn−1+1)

nn−1 nn−1 nn−1 nn−1 (x −1 ) = [σ(x −1 )] ⇔ [x −1 ] = (σ(x −1 )) in in−1=(nn−2+1) in in−1=(nn−2+1) in in−1=(nn−2+1) in in−1=(nn−2+1) ......

n2 n2 n2 n2 (xi2 )i2=(n1+1) = [σ(xi2 )]i2 =(n1+1) ⇔ [xi2 ]i2=(n1+1) = (σ(xi2 ))i2 =(n1+1)

n1 n1 n1 n1 (xi1 )i1=1 = [σ(xi1 )]i1=1 ⇔ [xi1 ]i1=1 = (σ(xi1 ))i1 =1 Therefore, since

x1 · · · xi1 · · · xn1 · · · xn −1+1 · · · xj · · · xn σ =  n k n  ,  σ(x1) · · · σ(xi1 ) · · · σ(xn1 ) · · · σ(xnn−1+1) · · · σ(xjk ) · · · σ(xnn )  then

− σ(x1) · · · σ(xi1 ) · · · σ(xn1 ) · · · σ(xnn−1+1) · · · σ(xjk ) · · · σ(xnn ) σ 1 =   ,  x1 · · · xi1 · · · xn1 · · · xnn−1+1 · · · xjk · · · xnn  so

−1 n1 n2 n3 nn−1 N (σ ) = (σ(x 1 )) (σ(x 2 )) (σ(x 3 )) · · · (σ(x −1 )) λ i i1 =1 i i2 =(n1+1) i i3=(n2+1) in in−1=(nn−2+1)

nn nn nn−1 n3 n2 n1 (σ(x )) [x ] [x −1 ] · · · [x 3 ] [x 2 ] [x 1 ] in in=(nn−1+1) in in=(nn−1+1) in in−1=(nn−2+1) i i3=(n2+1) i i2=(n1+1) i i1=1 −1 is a GSP. Hence, σ ∈ GSP Pλ(Sn). The converse can be proved in a similar way since (σ−1)−1 = σ.

2. Also, σ ∈ GSP Pρ(Sn) implies

n1 n2 n3 nn−1 nn Nρ(σ) = (xi1 ) (xi2 ) (xi3 ) · · · (xi −1 ) (xi ) i1=1 i2 =(n1+1) i3=(n2+1) n in−1=(nn−2+1) n in=(nn−1+1)

n1 n2 n3 nn−1 nn (σ(xi1 )) (σ(xi2 )) (σ(xi3 )) · · · (σ(xi −1 )) (σ(xi )) i1=1 i2 =(n1+1) i3=(n2+1) n in−1=(nn−2+1) n in=(nn−1+1) N N N is a GSP, where xij ∈ , ∀ ij ∈ , j ∈ and nn = n. The interval [1, n] is partitioned into

[1, n] = [1, n1] ∪ [n1 + 1, n2] ∪ · · · ∪ [nn−2 + 1, nn−1] ∪ [nn−1, nn].

nj The length of each grouping (·)ij is determined by the corresponding interval of integers [ni + 1, ni+1] and it is a matter of choice in other to make the number Nρ(σ) a GSP. 76 Tèm´ıtópé Gbóláhàn Ja´ıyéo.lá No. 3

Now that Nρ(σ) is a GSP, the following are true:

nn n1 (x ) = (σ(x 1 )) in in=(nn−1+1) i i1 =1

nn−1 n2 (x −1 ) = (σ(x 2 )) in in−1=(nn−2+1) i i2 =(n1+1) ......

n2 nn−1 (x 2 ) = (σ(x −1 )) i i2=(n1+1) in in−1=(nn−2+1)

n1 nn (x 1 ) = (σ(x )) . i i1=1 in in=(nn−1+1) Therefore, since

− σ(x1) · · · σ(xi1 ) · · · σ(xn1 ) · · · σ(xnn−1+1) · · · σ(xj ) · · · σ(xnn ) σ 1 =  k  .  x1 · · · xi1 · · · xn1 · · · xnn−1+1 · · · xjk · · · xnn 

−1 n1 n2 n3 nn−1 N (σ ) = (σ(x 1 )) (σ(x 2 )) (σ(x 3 )) · · · (σ(x −1 )) ρ i i1=1 i i2=(n1+1) i i3 =(n2+1) in in−1=(nn−2+1)

nn n1 n2 n3 nn−1 nn (σ(x )) (x 1 ) (x 2 ) (x 3 ) · · · (x −1 ) (x ) in in=(nn−1+1) i i1 =1 i i2=(n1+1) i i3=(n2+1) in in−1=(nn−2+1) in in=(nn−1+1) −1 is a GSP. Hence, σ ∈ GSP Pρ(Sn). The converse can be proved in a similar way since (σ−1)−1 = σ. 3. 1 2 · · · n I =   .  1 2 · · · n 

Nλ(I) = 12 · · · nn · · · 21 = 12 · · · (nn) · · · 21 =⇒ I ∈ GSP Pλ(Sn) and

Nρ(I) = (12 · · · n)(12 · · · n) =⇒ I ∈ GSP Pρ(Sn) thus I ∈ GSP P (Sn).

§4. Conclusion and Future studies

By Theorem 3.1, it is certainly true in every symmetric group Sn of degree n there exist at least a RGSPP and a LGSPP(although they are actually RPP and LPP). Following Example

3.1, there are 2 RGSPPs, 2 LGSPPs and 2 GSPPs in S2 while from Lemma 3.1, there are 4

RGSPPs, 4 LGSPPs and 2 GSPPs in S3. Also, it can be observed that

|GSP Pρ(S2)| + |GSP Pλ(S2)| − |GSP P (S2)| = 2! = |S2| Vol. 2 Palindromic permutations and generalized Smarandache palindromic permutations 77 and

|GSP Pρ(S3)| + |GSP Pλ(S3)| − |GSP P (S3)| = 3! = |S3|. The following problems are open for further studies.

Problem 4.1 1. How many RGSPPs, LGSPPs and GSPPs are in Sn?

|GSP Pρ(Sn)| + |GSP Pλ(Sn)| − |GSP P (Sn)| = n! = |Sn|? hold. If not, for what other n > 3 is it true? The GAP package or any other appropriate mathematical package could be helpful in investigating the solutions to them. If the ﬁrst question is answered, then the number of palindromes that can be formed from the set {1, 2, · · · , n} can be known since in the elements of Sn, the bottom row gives all possible permutation of the integers 1, 2, · · · , n. The Cayley Theorem(Theorem 2.1) can also be used to make a further study on generalized Smarandache palindromic permutations. In this work, N was the focus and it does not contain the integer zero. This weakness can be strengthened by considering the set Zn = {0, 1, 2, · · · , n−

1}, ∀ n ∈ N. Recall that (Zn, +) is a group and so by Theorem 2.1 (Zn, +) is isomorphic to a permutation group particularly, one can consider a subgroup of the symmetric group SZn .

References

[1] C. Ashbacher, L. Neirynck, The density of generalized Smarandache palindromes, http://www.gallup.unm.edu/˜ smarandache/GeneralizedPalindromes.htm. [2] G. Gregory, Generalized Smarandache palindromes, http://www. gallup. unm. edu/˜ smarandache/GSP. htm. [3] C. Hu (2000), On Smarandache generalized palindrome, Second International Con- ference on Smarandache Type Notions In Mathematics and Quantum Physics, University of Craiova, Craiova Romania, atlas-conferences.com/c/a/f/t/25.htm. [4] M.Khoshnevisan, Manuscript, 2003. [5] M. Khoshnevisan, Generalized Smarandache palindrome, Mathematics Magazine, Au- rora, Canada, 2003. [6] M. Khoshnevisan, Proposed problem 1062, The PME Journal, USA, 11(2003), pp. 501. [7] K. Ramsharan, Manusript, 2003. www.research.att.com/˜ njas/sequences/A082461. [8] F. Smarandache, Sequences of Numbers Involved in unsolved problems, http://www. gallup. unm. edu/˜ smarandache/eBooks-otherformats. htm, 2006, pp. 141. Scientia Magna Vol. 2 (2006), No. 3, 78-80

On certain inequalities involving the Smarandache function

J´ozsefS´andor

Department of Mathematics and Computer Sciences, Babe¸s-Bolyai University of Cluj, Cluj, Romania

Abstract Let S(n) be the Smarandache function deﬁned by S(n) = min{m ≥ 1 : n|m!}. The aim of this note is to prove that for each k ≥ 2 there are inﬁnitely many positive integers

m1, . . . , mk and n1, . . . , nk such that S(m1 + m2 + ··· + mk) < S(m1) + S(m2) + ··· + S(mk),

and S(n1 + n2 + ··· + nk) > S(n1) + S(n2) + ··· + S(nk). Keywords Vinogradov’s three-primes theorem, inequalities. §1. Introduction

For any positive integer n, the famous Smarandache function is deﬁned by

S(n) = min{m ≥ 1 : n|m!}

For many properties, problems, generalizations and extensions of this function see e.g. Smaran- dache’s well-known book [5], or the author’s book [3]. Recently, Lu Yaming [6] has proved that for each positive integer k, the equation

S(x1 + x2 + ··· + xk) = S(x1) + S(x2) + ··· + S(xk)

has inﬁnitely many positive integer solutions x1, . . . , xk. In what follows, we shall prove that both of inequalities S(a + b) < S(a) + S(b) and S(c + d) > S(c) + S(d) have inﬁnitely many positive integer solutions a, b, c, d; and more generally the same is true for S(m1 + ··· + mk) <

S(m1) + ··· + S(mk), and S(n1 + ··· + nk) > S(n1) + ··· + S(nk), for each k ≥ 2.

§2. Proof of the theorem

The fact that S(a + b) < S(a) + S(b) has inﬁnitely many solutions, follows from well known inequalities. For example, a = n! and b = (n + 1)! are solutions. Indeed, as a + b = n! + (n + 1)! = n!(n + 2), and n!(n + 2) dividing (n + 2)!, one has clearly S(a + b) ≤ n + 2 < 2n + 1 = S(n!) + S((n + 1)!), as S(n!) = n, etc. For another solutions, put a = p! + 1, b = p! − 1, where p ≥ 2 is a prime. Then, it is known that (see e.g. [2], [3]) S(p! + 1) > p, S(p! − 1) > p, so we get S(p! + 1) + S(p! − 1) > 2p ≥ S(2p!), as S(2p!) ≤ 2S(p!) = 2p, by the inequality S(mn) ≤ nS(m), see [1], [3]. Vol. 2 On certain inequalities involving the Smarandache function 79

Let now c, d be positive integers such that c + d is a prime, and c > 4, d > 4. Then c and d cannot be both primes, since then c+d would be even (> 4). Then S(c+d) = c+d > S(c)+S(d), by the well-known fact (see e.g. [3], [5]) that S(n) ≤ n, with equality only for n = 4, and n = prime. The last part of the above proof may be easily extended to the case of k numbers. Namely,

let n1, n2, . . . , nk be positive integers, not all of which are primes, so that n1 + n2 + ··· + nk is a

prime. (It is easy to see the possibility of constructing such numbers). Then S(n1 + ··· + nk) =

n1 + ··· + nk > S(n1) + ··· + S(nk), as above. For the extension of the ﬁrst part to arbitrary k, however, we need another argument. As in [6], we need Vinogradov’s three-primes theorem, as stated in Lemma 1, and its generalization, as stated in Lemma 2: Lemma 1. There exists a constant K > 0, such that each odd integer n > K can be

written as n = p1 + p2 + p3, where pi(i = 1, 3) are odd primes. Lemma 2. If k ≥ 3 is an odd integer, then any suﬃciently large odd integer n can be written as a sum of k odd primes

n = p1 + p2 + ··· + pk.

This follows immediately from Lemma 1, by induction see [6]. s Let now k ≥ 3 be odd. Then the odd integer n = 4 − 1 for suﬃciently large s ≥ s0 may be written as s 4 − 1 = p1 + ··· + pk.

On the other hand, S(4s − 1) < 4s − 1, as 4s − 1 ≡ 0 (mod 3), so 4s − 1 6= prime, 6= 4. This implies

s S(p1 + ··· + pk) = S(4 − 1) < p1 + ··· + pk = S(p1) + ··· + S(pk).

If k ≥ 4 is even, then for suﬃciently large s, 3s − 2 may be written as

s 3 − 2 = p1 + ··· + pk−1

implying s 3 = 2 + p1 + ··· + pk−1,

s s s i.e. S(3 ) = S(2 + p1 + ··· + pk−1) < 3 = 2 + p1 + ··· + pk1 = S(2) + S(p1) + ··· + S(pk−1), 3 not being a prime. This ﬁnishes the proof of the theorem. Remark. The arithmetical function occurring in the ﬁrst part of the proof, namely

F (n) = min{m ≥ 0 : n|m!(m + 2)}

(where 0! = 1) has a close analogy with the Smarandache function. We have called it in [4], as the modification of the Smarandache function. In [4] we have proved that F (p) = p − 1 for all primes p; F (k!) = k + 1 for k ≥ 3; F (1!) = F (2!) = 1; and that S(n) < F (n) or S(m) > F (m) both have infinitely many solutions n, m. 80 JózsefSándor No. 3 References

[1] J. Sándor,On certain inequalities involving the Smarandache function, Smarandache Notions Journal, 7(1996), pp. 3-6. [2] J. Sándor,On certain new inequalities and limits for the Smarandache function, Smaran- dache Notions Journal, 9(1998), No.1-2, pp. 63-69. [3] J. Sándor,Geometric theorems, diophantine equations, and arithmetic functions, Amer- ican Research Press, New Mexico, 2002. [4] J. Sándor,On a modification of the Smarandache function, Octogon Math. Mag. 12(2004), No.2A, pp. 521-523. [5] F. Smarandache, Only Problems, Xiquan Publ. House, Chicago, 1993. [6] Lu Yaming, On the solutions of an equation involving the Smarandache function, Sci- entia Magna, 2(2006), No.1, pp. 76-79. Scientia Magna Vol. 2 (2006), No. 3, 81-87

Sequences of pentagonal numbers1

Arun S. Muktibodh Mohota Science College, Umred Rd. Nagpur-440009, India.

Abstract Shyam Sunder Gupta [3] has defined Smarandache consecutive and reversed Smarandache sequences of triangular numbers. Delfim F.M.Torres and Viorica Teca [1] have further investigated these sequences and defined mirror and symmetric Smarandache sequences of triangular numbers making use of Maple system. Working on the same lines we have defined and investigated consecutive, reversed, mirror and symmetric Smarandache sequences of pentagonal numbers of dimension 2 using the Maple system . Keywords Figurate number, Smarandache consecutive sequence, Smarandache mirror sequence.

§1. Introduction

Figurate number is a number which can be represented by a regular geometrical arrangement of equally spaced points. If the arrangement forms a regular polygon the number is called a polygonal number. Different figurate sequences are formed depending upon the dimension we consider. Each dimension gives rise to a system of figurate sequences which are infinite in number.

In this paper we consider a ﬁgurate sequence of pentagonal numbers of dimension 2. Hence-

forth, unless and otherwise stated, pentagonal numbers ¡ will mean pentagonal numbers of dimension 2. 5 2 The n-th pentagonal number tn, n ∈ N of dimension 2 is deﬁned by tn = 2n+ 2 n(n−1)−n . We can obtain the ﬁrst k terms of Pentagonal numbers in Maple as: > t:= n->2*n + (5\2)*n*(n-1)-n^{2}:

> first := k -> seq(t(n), n=1...20); For example ﬁrst 20 terms are: > first(20);

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590 For constructing Smarandache sequence of pentagonal numbers we require the operation of concatenation on the terms of above sequence which is deﬁned in Maple as;

1This work is supported by UGC under Minor project F. No. 23-245/06. 82 Arun S. Muktibodh No. 3

> conc :=(n,m) -> n*10^length(m)+m:

We deﬁne Smarandache consecutive sequence {scsn} for pentagonal numbers recursively as;

scs1 = u1,

scsn = conc(scsn−1, un) Using Maple We have obtained ﬁrst 20 terms of Smarandache consecutive sequence of pentagonal numbers;

> conc :=(n,m)-> n*10^length(m)+m: > scs_n := (u,n)-> if n = 1 then u(1)else conc(scs_n(u,n-1),u(n)) fi: > scs := (u,n)-> seq (scs_n(u,i),i=1...n): > scs(t,20);

1, 15, 1512, 151222, 15122235, 1512223551, 151222355170, 15122235517092, 15122235517092117, 15122235517092117145, 15122235517092117145176, 15122235517092117145176210, 15122235517092117145176210247, 15122235517092117145176210247287, 15122235517092117145176210247287330, 15122235517092117145176210247287330376, 15122235517092117145176210247287330376425, 15122235517092117145176210247287330376425477, 15122235517092117145176210247287330376425477532,

15122235517092117145176210247287330376425477532590

Same sequence can be displayed in triangular form ¡ as;

> show := L -> map(i ->print(i),L): > show([scs(t,20)]);

1 15 1512 151222 15122235 1512223551 151222355170 15122235517092 15122235517092117 15122235517092117145 15122235517092117145176 15122235517092117145176210 15122235517092117145176210247 15122235517092117145176210247287 Vol. 2 Sequences of pentagonal numbers 83

15122235517092117145176210247287330 15122235517092117145176210247287330376 15122235517092117145176210247287330376425 15122235517092117145176210247287330376425477 15122235517092117145176210247287330376425477532 15122235517092117145176210247287330376425477532590

The reversed Smarandache sequence (rss)associated with a given sequence {un}, n ∈ N is de- ﬁned recursively as;

rss1 = u1,

rssn = conc(un, rssn−1).

In Maple we use the following program;

> rss_n :=(u,n) -> if n=1 then u(1) else conc(u(n),rss_n(u,n-1)) fi: > rss := (u,n) -> seq(rss_n(u,i),i=1..n):

We get the ﬁrst 20 terms of reversed Smarandache sequence of pentagonal numbers;

> rss(t,20);

1, 51, 1251, 221251, 35221251, 5135221251, 705135221251, 92705135221251, 11792705135221251, 14511792705135221251, 17614511792705135221251, 21017614511792705135221251, 24721017614511792705135221251, 28724721017614511792705135221251, 33028724721017614511792705135221251, 37633028724721017614511792705135221251, 42537633028724721017614511792705135221251, 47742537633028724721017614511792705135221251, 53247742537633028724721017614511792705135221251, 59053247742537633028724721017614511792705135221251

Smarandache Mirror Sequence (sms)is deﬁned as follows ¢

sms1 = u1,

smsn = conc(conc(un, smsn−1), un).

The following program gives ﬁrst 20 terms of Smarandache Mirror sequence of pentagonal numbers.

> sms_n := (u,n) -> if n=1 then 84 Arun S. Muktibodh No. 3

> u(1) > else > conc(conc(u(n),sms_n(u,n-1)),u(n)) > fi: > sms :=(u,n) ->seq(sms_n(u,i), i=1..n): > sms(t,20);

1, 515, 1251512, 22125151222, 352212515122235, 5135221251512223551, 70513522125151222355170, 927051352212515122235517092, 117927051352212515122235517092117, 145117927051352212515122235517092117145, 176145117927051352212515122235517092117145176, 210176145117927051352212515122235517092117145176210, 247210176145117927051352212515122235517092117145176210247, 2872472101761451179270513522125151222355170921171451762102\ 47287, 3302872472101761451179270513522125151222355170921171\ 45176210247287330, 3763302872472101761451179270513522125151\ 22235517092117145176210247287330376, 4253763302872472101761\ 4511792705135221251512223551709211714517621024728733037642\ 5, 47742537633028724721017614511792705135221251512223551709\ 2117145176210247287330376425477, 53247742537633028724721017\ 6145117927051352212515122235517092117145176210247287330376\ 425477532, 590532477425376330287247210176145117927051352212\ 515122235517092117145176210247287330376425477532590.

Finally Smarandache Symmetric sequence (sss) is deﬁned as:

sss2n−1 = conc(bld(scs2n−1), rss2n−1),

sss2n = conc(scs2n, rss2n), n ∈ N.

where the function bld ¡ (But Last Digit) is deﬁned in Maple as:

> bld := n->iquo(n,10):

First 20 terms of Smarandache Symmetric sequence are obtained in Maple as:

> bld :=n-> iquo(n,10): > conc := (n,m)-> n*10^length(m)+m: > sss_n := (u,n) -> if type(n,odd) then > conc(bld(scs_n(u,(n+1)/2)),rss_n(u,(n+1)/2)) > else > conc(scs_n(u,n/2),rss_n(u,n/2)) > fi: > sss := (u,n) -> seq(sss_n(u,i), i=1..n): > sss(t,20); Vol. 2 Sequences of pentagonal numbers 85

1, 11, 151, 1551, 1511251, 15121251, 15122221251, 151222221251, 151222335221251, 1512223535221251, 1512223555135221251, 15122235515135221251, 15122235517705135221251, 151222355170705135221251, 151222355170992705135221251, 1512223551709292705135221251, 151222355170921111792705135221251, 1512223551709211711792705135221251, 151222355170921171414511792705135221251, 1512223551709211714514511792705135221251.

We find out primes from a large number (first 500 terms) of various Smarandache sequences defined so far. We have used Maple 6 on Pentium 3 with 256 Mb RAM. We first collect the lists of first 500 terms of the consecutive, reversed, mirror and symmetric sequences of Pentagonal numbers:

> st :=time(): Lscs500:=[scs(t,500)]: printf("%a seconds",round(time()-st)); 16 seconds > st :=time(): Lrss500:=[rss(t,500)]: printf("%a seconds",round(time()-st));

18 seconds > st :=time(): Lsms500:=[sms(t,500)]: printf("%a seconds",round(time()-st)); 50 seconds > st :=time(): Lsss500:=[sss(t,500)]: printf("%a seconds",round(time()-st)); 11 seconds

Further we ﬁnd the number of digits in the 500th term of each sequence:

> length(Lscs500[500]),length(Lrss500[500]);

2626, 2626

> length(Lsms500[500]),length(Lsss500[500]);

5251, 2268

There exist no prime in the ﬁrst 500 terms of Smarandache consecutive sequence of pentagonal numbers.

>st:= time():select(isprime,Lscs500); > printf("%a minutes",round((time()-st)/60));

[ ] 86 Arun S. Muktibodh No. 3

11 minutes

There are only two primes in the ﬁrst 500 terms of reversed Smarandache sequence of pentagonal numbers.

>st:= time():select(isprime,Lrss500); [221251, 10049299717989459817697410966479588795130943769362592877921329139 09065189915891828845287725870018628085562848478413583426827208201 78131780620799267923578547778627718076501758257515274482738157315 172490718327117770525698766923068587679476731066676660456541764792 641706355162935623226171261105605015990059302587075811557526569405 635755777552005462654055534875292252360518015124550692501424959549 051485104797247437469054637645850453274480744290437764326542757422 524175041251407554026239772392853880138320378423736736895364263596 0354973503734580341263367533227327823234031901314653103230602301752 9751293302891228497280852767627270268672646726070256762528524897245 1224130237512337523002226322226521901215402118220827204752012619780 1943719097187601842618095177671744217120168011648516172158621555515 2511495014652143571406513776134901320712927126501237612105118371157 2113101105110795105421029210045980195609322908788558626840081777957 7740752673157107690267006501630561125922573555515370519250174845467 6451043474187403038763725357734323290315130152882275226252501238022 6221472035192618201717161715201426133512471162108010019258527827156 5159053247742537633028724721017614511792705135221251]

> printf("%a minutes",round((time()-st)/60)); 45 minutes

There is no prime in the ﬁrst 500 terms of Smarandache mirror sequence.

>st:= time():select(isprime,Lsms500); > printf("%a minutes",round((time()-st)/60));

[ ]

312 minutes

There are 4 primes in ﬁrst 500 terms of Smarandache symmetric sequence of pentagonal numbers.

> st:= time(): Vol. 2 Sequences of pentagonal numbers 87

> select(isprime,Lsss500); > printf("%a minutes",round((time()-st)/60));

[11, 151, 1512223551709211714514511792705135221251, 1512223551709\ 2117145176210247287330376425477532590651715782852925100100\ 1925852782715651590532477425376330287247210176145117927051\ 35221251]

50 minutes

There are some results which can be obtained by the readers as; How many pentagonal numbers are there in Smarandache consecutive , mirror and symmetric sequences of pentagonal numbers ?

References

[1] Delﬁm F.M., Viorica Teca, Consective, Reversed, Mirror and Symmetric Smarandache Sequences of Triangular Numbers, Scientia Magna, 2(2005), pp. 39-45. [2] Muktibodh A.S., Figurate Systems, Bull. Marathwada Mathematical Soc., 2(2005), pp. 12-20. [3] Shyam Sunder Gupta, Smarandache Sequence of Triangular Numbers, Smarandache Notions Journal, 14, pp. 366-368. Scientia Magna Vol. 2 (2006), No. 3, 88-91

On the additive analogues of the simple function

Jianli Gao

Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract The main purpose of this paper is using the analytic method to study the convergent properties of the Jozse Sandor’s function, and obtain some interesting results. Keywords Riemann zeta-function, convergent, asymptotic formula.

§1. Introduction and conclusions

For any positive integer n, the function Z∗(n) and Z(n) will be deﬁned as ½ ¾ m(m + 1) Z (n) = max m ∈ N : ≤ n , ∗ 2

and ½ ¾ m(m + 1) Z(n) = min m ∈ N : n ≤ . 2

m(m+1) That is, Z∗(n) denotes the greatest positive integer m such that 2 ≤ n, and Z(n) m(m+1) denotes the smallest positive integer m such that n ≤ 2 . For example: Z∗(1) = 1, Z∗(2) = 1, Z∗(3) = 2, Z∗(4) = 2, Z∗(5) = 2, Z∗(6) = 3, Z∗(7) = 3,

Z∗(8) = 3, Z∗(9) = 3, ··· and Z(1) = 1, Z(2) = 2, Z(3) = 2, Z(4) = 3, Z(5) = 3, Z(6) = 3, Z(7) = 4, Z(8) = 4, Z(9) = 4, Z(10) = 4, ··· . Recently, Jozsef Sandor had studied the properties of these functions, and proved some interesting conclusions, one of them is

X∞ 1 (Z (n))s n=1 ∗

is convergent for s > 2, and divergent for s ≤ 2. In this paper, we shall use the elementary methods to study the convergent properties of some series involving the Jozsef Sandor’s functions, and obtain some interesting identities. That is, we shall prove the following several theorems: Theorem 1. For any complex number s, the inﬁnite series

X∞ (−1)n , (Z (n))s n=1 ∗ Vol. 2 Jianli Gao 89 and X∞ (−1)n , (Z(n))s n=1 are convergent if s > 0, and divergent if s ≤ 0. Theorem 2. For any complex number s with Re s > 2, we have the identities

X∞ 1 = ζ(s − 1) + ζ(s), (Z (n))s n=1 ∗ and X∞ 1 = ζ(s − 1), (Z(n))s n=1 where ζ(s) denotes the Riemann zeta-function. Theorem 3. For any positive integer n and complex number s with Re s > 1, we have

X∞ (−1)n 2 1 = ζ(s) − ζ(s), (Z (n))s 4s 2s n=1 ∗ and µ ¶ µ ¶ X∞ (−1)n 1 2 1 = 1 − ζ(s) − ζ s, , (Z(n))s 2s 4s 4 n=1 where ζ(s, α) denotes the Hurwitz zeta-function. From Theorem 3 we may immediately deduce the following:

Corollary. Let Z∗(n) deﬁned as the above, then we have

X∞ (−1)n π2 = − . (Z (n))2 16 n=1 ∗ §2. Proof of the theorems

In this section, we shall complete the proof of Theorems. First we prove Theorem 1. If m(m + 1) (m + 1)(m + 2) ≤ n < , 2 2 then Z∗(n) = m repeated (m + 1)(m + 2) m(m + 1) − = m + 1 2 2 times, so we have X∞ (−1)n X∞ (−1)n(m + 1) = (Z (n))s ms n=1 ∗ n=1 Z∗(n)=m If m is an odd, every term is concealed by positive and negative addition. If m is an even, only one term will be retained in every repeated term, so we can obtain µ ¶ X∞ (−1)n 1 1 1 (−1)n 1 1 1 1 = − + − + ··· + + ··· = − − + + ··· . (Z (n))s 2s 4s 6s (2n)s 2s 1s 2s 3s n=1 ∗ 90 On the additive analogues of the simple function No. 3

Using the same method, we can also obtain

X∞ (−1)n 1 1 1 1 (−1)n = − + − + + ··· + + ··· . (Z(n))s 1s 3s 5s 7s (2n − 1)s n=1

Now Theorem 1 follows from the Leibniz convergent criterion. If m(m + 1) (m + 1)(m + 2) ≤ n < , 2 2 then Z∗(n) repeated (m + 1)(m + 2) m(m + 1) − = m + 1 2 2 times. Hence, we have

X∞ 1 X∞ m + 1 = (Z (n))s ms n=1 ∗ m=1 = ζ(s − 1) + ζ(s).

Using the same method, we can also obtain

X∞ 1 X∞ m(m+1) − (m−1)m = 2 2 (Z(n))s ms n=1 m=1 = ζ(s − 1).

This completes the proof of Theorem 2. Now we come to prove Theorem 3. From the method of proving Theorem 1 and Theorem 2, we can easily get

X∞ (−1)n 1 1 1 (−1)n s = − s + s − s + ··· + s + ··· (Z∗(n)) 2 4 6 (2n) n=1 µ ¶ 1 1 1 1 = − + + + ··· + + ··· 2s 6s 10s (4n − 2)s µ ¶ 1 1 1 + + + ··· + + ··· 4s 8s (4n)s µ ¶ 1 1 1 1 = − + + ··· + + ··· 2s 1s 3s (2n − 1)s µ ¶ 1 1 1 1 + + + ··· + + ··· 2s 2s 4s (2n)s 2 1 = ζ(s) − ζ(s), 4s 2s Vol. 2 Jianli Gao 91 and

X∞ (−1)n 1 1 1 1 (−1)n = − + − + + ··· + + ··· (Z(n))s 1s 3s 5s 7s (2n − 1)s n=1 µ ¶ 1 1 1 1 = + + + ··· + + ··· 1s 3s 5s (2n − 1)s µ ¶ 1 1 1 −2 + + ··· + + ··· 1s 5s (4n − 3)s µ ¶ µ ¶ 1 2 1 = 1 − ζ(s) − ζ s, , 2s 4s 4 where we have used the identities µ ¶ X∞ 1 1 = 1 − ζ(s), (2n − 1)s 2s n=1 and µ ¶ 1 1 1 X∞ 1 1 X∞ 1 1 1 + + ··· + + ··· = = ¡ ¢ = ζ s, . 1s 5s (4n − 3)s (4n + 1)s 4s 1 s 4s 4 n=0 n=0 n + 4 This completes the proof of Theorem 3.

References

[1] Jozsef Sandor, On additive analogues of certain arithmetic function, Samarandache Notions Journal, 14(2004), pp. 128-134. [2] Jian Zhao and Chao Li, On the mean value of new arithmetical function, Research on Smarandache Problems in Number Theory, Hexis., 2005, pp. 6-40. Scientia Magna Vol. 2 (2006), No. 3, 92-95

On the primitive numbers of power p

Weiyi Zhu

College of Mathematics, Physics and Information Engineer Zhejiang Normal University, Jinhua, Zhejiang, P.R.China

Abstract For any positive integer n and prime p, let Sp(n) denotes the smallest positive integer m such that m! is divisible by pn. The main purpose of this paper is using the X elementary method to study the properties of the summation Sp(d), and give an exact d|n calculating formula for it. Keywords Primitive number of power p, summation, calculating formula.

§1. Introduction and Results

Let p be a prime and n be any positive integer. Then we deﬁne the primitive number n function Sp(n) of power p as the smallest positive integer m such that m! is divisible by p .

For example, S3(1) = 3,S3(2) = 6,S3(3) = S3(4) = 9, ······ . In problem 49 of book [1],

Professor F.Smarandache asked us to study the properties of the sequence {Sp(n)}. About this

problem, Zhang Wenpeng and Liu Duansen [3] had studied the asymptotic properties of Sp(n), and obtained an interesting asymptotic formula for it. That is, for any ﬁxed prime p and any positive integer n, they proved that µ ¶ p S (n) = (p − 1)n + O ln n . p ln p

Yi Yuan [4] had studied the asymptotic property of Sp(n) in the form 1 X |S (n + 1) − S (n)| , p p p n≤x

and obtained the following conclusion: For any real number x ≥ 2, we have µ ¶ µ ¶ 1 X 1 ln x |S (n + 1) − S (n)| = x 1 − + O . p p p p ln p n≤x

Xu Zhefeng [5] had studied the relationship between the Riemann zeta-function and an

inﬁnite series involving Sp(n), and obtained some interesting identities and asymptotic formulae

for Sp(n). That is, for any prime p and complex number s with Res > 1, we have the identity:

X∞ 1 ζ(s) = , Ss(n) ps − 1 n=1 p Vol. 2 On the primitive numbers of power p 93

where ζ(s) is the Riemann zeta-function. And, let p be a ﬁxed prime, then for any real number x ≥ 1,

X∞ µ ¶ 1 1 p ln p − 1 +ε = ln x + γ + + O(x 2 ), S (n) p − 1 p − 1 n=1 p Sp(n)≤x

where γ is the Euler constant, ε denotes any ﬁxed positive number. Chen Guohui [7] had studied the calculating problem of the special value of the Smaran- dache function S(n) = min{m : m ∈ N, n|m!}. That is, let p be a prime and k an integer with

n n n1 1 ≤ k < p. Then for polynomials f(x) = x k +x k−1 +···+x with nk > nk−1 > ··· > n1 ≥ 1, we have: S(pf(p)) = (p − 1)f(p) + pf(1).

And, let p be a prime and k an integer with 1 ≤ k < p. Then for any positive integer n, we have: ³ ´ µ ¶ n 1 S pkp = k φ(pn) + p, k

where φ(n) is the Euler function. All these conclusions also hold for primitive function Sp(n) of power p. In this paper, we shall use the elementary method to study the calculating problem of the summation X Sp(d), d|n and give some interesting calculating formulas for it. That is, we shall prove the following conclusions: Theorem. For any prime p, we have the calculating formulas X (1) Sp(d) = pσ(n), if 1 ≤ n ≤ p; Xd|n (2) Sp(d) = pσ(n) − (n − 1)p, if p < n ≤ 2p, where σ(n) denotes the summation over d|n all divisors of n. X For general positive integer n > 2p, whether there exists a calculating formula for Sp(d) d|n is an open problem.

§2. Proof of Theorem

To complete the proof of the theorem, we need a simple lemma which stated as following: Lemma. For any prime p, we have:

(1) Sp(d) = dp, if 1 ≤ d ≤ p;

(2) Sp(d) = (d − k + 1)p, if (k − 1)p + k − 2 < d ≤ kp. n Proof. First we prove the case (1). From the deﬁnition of Sp(n) = min{m : p |m!} we know that to prove the case (1) of Lemma, we only to prove that pdk(dp)!. That is, pd|(dp)! and 94 Weiyi Zhu No. 3

· ¸ X∞ dp pd+1 † (dp)!. According to Theorem 1.7.2 of reference [6], we only to prove that = d. pj · ¸ j=1 d In fact, if 1 ≤ d < p, note that = 0 (j = 1, 2, ······ ), we have pj · ¸ · ¸ · ¸ X∞ dp d d = d + + + ··· = d. pj p p2 j=1 · ¸ X dp That is means S (d) = dp. If d = p, then = d + 1, but pp † (p2 − 1)! and pp|p2!. So p pj d|n 2 from the deﬁnition of Sp(n) we have Sp(p) = p = dp. This proves the case (1) of Lemma. Now we prove the case (2) of Lemma. Using the same· method¸ of proving· ¸ the case (1) d − 1 d − 1 of Lemma we can also deduce that if p < d ≤ 2p, then = 1, = 0, (j = pj pj 2, 3, ······ ), we have · ¸ · ¸ · ¸ X∞ (d − 1)p d − 1 d − 1 = d − 1 + + + ··· = d. pj p p2 j=1

That is means that Sp(d) = (d − 1)p. From Theorem 1.7.2 of reference [6] we know that if d p < d ≤ 2p, then p k((d − 1)p)!. That is, Sp(d) = (d − 1)p. This proves the lemma. Now we use this Lemma to complete the proof of the theorem. First we prove the case (1) of Theorem. From the case (1) of Lemma we know that if 1 ≤ n ≤ p, then X X X Sp(d) = dp = p d = pσ(n). d|n d|n d|n Now we prove the case (2) of Theorem. We ﬁnd that if p < n ≤ 2p and d|n, then there is only one divisor d(> p) of n, i.e. d = n, the reason is that if d > p and d|n, then n = d, 2d, 3d, ··· , but 2d > 2p, ··· . So we may immediately deduce the following conclusion: if p < n ≤ 2p, then X X X X Sp(d) = Sp(d) + Sp(d) = p d + Sp(n) = pσ(n) + (n − 1)p. d|n d|n d|n d|n 1≤d≤p p

This completes the proof of Theorem.

References

[1] F. Smarandache. Only Problems, Not Solutions. Chicago, Xiquan Publishing House, 1993. [2] Tom M. Apostol. Introduction to Analytic Number Theory. New York, Springer-Verlag, 1976. [3] Zhang Wenpeng and Liu Duansen. On the primitive numbers of power p and its asymptotic property. Smarandache Notions Journal, Vol. 13(2002), 173-175. Vol. 2 On the primitive numbers of power p 95

[4] Yi Yuan. On the primitive numbers of power p and its asymptotic property. Scientia Magna, Vol. 1(2005), No. 1, 175-177. [5] Xu Zhefeng. Some arithmetical properties of primitive numbers of power p. Scientia Magna, Vol. 2(2006), No. 1, 9-12. [6] Pan Chengdong and Pan Chengbiao. The Elementary Number Theory. Beijing Uni- versity Press, Beijing, 2003. [7] Chen Guohui. Some exact calculating formulas for the Smarandache function. Scientia Magna, Vol. 2(2006), No. 2, 95-97. Scientia Magna Vol. 2 (2006), No. 3, 96-103

Smarandache sums of products

Anant W. Vyawahare

Department of Mathematics, M. Mohota Science College, Nagpur University, NAGPUR-440009, India

Keywords The Smarandache sum, stirling numbers, properties of sequences.

Introduction This paper deals with the sums of products of first n natural numbers, taken r at a time. Many interesting results about the summations are obtained. Mr. Rama- subramanian [1] has already made some work in this direction. This paper is an extension of his work. In next part, the sums of odd and even natural numbers are discussed, and also of natural numbers, not necessarily beginning with one. After that, properties of sequences, arising out of these sums are obtained. Interestingly, the numbers thus obtained are Stirlings numbers. 1.1 Definition. The Smarandache sum of products is denoted by S(n, r), and is defined as sum of products of first n natural numbers, taken r at a time, without repeatition, r ≤ n. For example: S(4, 1) = 1 + 2 + 3 + 4 = 10, S(4, 2) = 1 · 2 + 1 · 3 + 1 · 4 + 2 · 3 + 2 · 4 + 3 · 4 = 35, S(4, 3) = 1 · 2 · 3 + 1 · 2 · 4 + 1 · 3 · 4 + 2 · 3 · 4 = 50, S(4, 4) = 1 · 2 · 3 · 4 = 24. We assume that S(n, 0) = 1. 1.2 Following are some elementary properties of S(n, r): 1. S(n, n) =∟n =factorial n = nS(n − 1, n − 1), 2. S(n, 1) = n(n + 1)/2 ; these are triangular numbers, 3. (p + 1)(p + 2)(p + 3) ... (p + n) = S(n, 0)pn + S(n, 1)pn−1 + S(n, 2)pn−2 + S(n, 3)pn−3 + ··· + S(n, n − 1)p + S(n, n), 4. S(n, 0) + S(n, 1) + S(n, 2) + ··· + S(n, n) = S(n + 1, n + 1) =∟(n + 1), 5. Number of terms in S(n, r) =n Cr. The 4th property can be obtained by putting p = 1 in the 3-rd property. Verification of 4th property for n = 5. S(5, 0) = 1, S(5, 1) = 1 + 2 + 3 + 4 + 5 = 15, S(5, 2) = 1 · 2 + 1 · 3 + 1 · 4 + 1 · 5 + 2 · 3 + 2 · 4 + 2 · 5 + 3 · 4 + 3 · 5 + 4 · 5 = 85, S(5, 3) = 1·2·3+1·2·4+1·2·5+1·3·4+1·3·5+2·3·4+2·3·5+3·4·5+2·4·5+2·4·5 = 225, S(5, 4) = 1 · 2 · 3 · 4 + 1 · 2 · 3 · 5 + 1 · 3 · 4 · 5 + 2 · 3 · 4 · 5 + 1 · 2 · 4 · 5 = 274, S(5, 5) = 1 · 2 · 3 · 4 · 5 = 120. Vol. 2 Smarandache sums of products 97

Hence, left side is 1 + 15 + 85 + 225 + 274 + 120 = 720,

right side is S(6, 6) = 1 · 2 · 3 · 4 · 5 · 6 = 720.

1.3 We have,

(p − 1)(p − 2)(p − 3) ... (p − n) = S(n, 0) − S(n, 1)pn + S(n, 2)pn−1 − S(n, 3)pn−3 + ... + S(n, n − 1)p + (−1)nS(n, n).

Put p = 1,

S(n, 0) − S(n, 1) + S(n, 2) − S(n, 3) + ··· + S(n, n − 1) + (−1)nS(n, n) = 0.

If n is even, then S(n, 0) + S(n, 2) + S(n, 4) + ... + S(n, n) = S(n, 1) + S(n, 3) + S(n, 5) + ... + S(n, n − 1), for odd n, S(n, 0) + S(n, 2) + S(n, 4) + ... + S(n, n − 1) = S(n, 1) + S(n, 3) + S(n, 5) + ... + S(n, n).

1.4 To verify S(n, r) = S(n − 1, r) + nS(n − 1, r − 1), r < n. Put r = 1, then, the left side is

S(n, 1) = n(n + 1)/2.

Right side is

S(n − 1, 1) + nS(n − 1, 0) = (n − 1)n/2 + n · 1 = n · (n + 1)/2.

2.1 Reduction formula (I) for S(n, r). Here we use the result of 1.4 repeatedly. We have, S(n, r) = S(n − 1, r) + nS(n − 1, r − 1),

S(n − 1, r) = S(n − 2, r) + (n − 1)S(n − 2, r − 1),

S(n − 2, r) = S(n − 3, r) + (n − 2)S(n − 3, r − 1),

S(n − 3, r) = S(n − 4, r) + (n − 1)S(n − 4, r − 1),

···

S(r + 1, r) = S(r, r) + (r + 1)S(r, r − 1).

Adding, we get,

S(n, r) = S(r, r) + (n)S(n − 1, r − 1) + (n − 1)S(n − 2, r − 1) + (n − 2)S(n − 3, r − 1) +... + (r + 1)S(r, r − 1)

Veriﬁcation: 98 Anant W. Vyawahare No. 3

Put n = 5, r = 2, the left side is

S(5, 2) = 85, right side is

S(2, 2) + 5S(4, 1) + 4S(3, 1) + 3S(2, 1) = 2 + 50 + 24 + 9 = 85,

2.2 Reduction formula (II) for S(n, r). We have,

(p + 1)(p + 2)(p + 3) ... (p + n)(p + n + 1) = S(n + 1, 0)pn+1 + S(n + 1, 1)pn + S(n + 1, 2)pn−1 + S(n + 1, 3)pn−2 + ... + S(n + 1, r + 1)pn−r + ... + S(n + 1, n + 1). (1)

Left side of (1) is

(p + 1){(p + 1 + 1)(p + 1 + 2)...(p + 1 + n − 1)(p + 1 + n)} = (p + 1){S(n, 0)(p + 1)n + S(n, 1)(p + 1)n−1 + ... + S(n, r)(p + 1)n−r ... + S(n, n)} = S(n, 0)(p + 1)n+1 + S(n, 1)(p + 1)n + ... + S(n, r)(p + 1)n−r+1 + ... + S(n, n).

Expanding each of (p + 1)n+1, (p + 1)n, (p + 1)n−1,..., (p + 1)n−r+1, by binomial theorem, we get the left side of (1) is

S(n, 0)[C(n + 1, 0)pn+1 + C(n + 1, 1)pn + ... + C(n + 1, r + 1)pn−r ... + C(n + 1, n + 1)] +S(n, 1){C(n, 0)pn + C(n, 1)pn−1 + ... + C(n, r)pn−r ... + C(n, n)} +S(n, 2){C(n − 1, 0)pn−1 + ... + C(n − 1, r − 1)pn−r ... + C(n + 1, n + 1)} + ... +S(n, r){C(n − r + 1, 1)pn−r + ...} +S(n, r + 1){C(n − r, 0)pn−r} + ... + S(n, n)(p + 1), (2)

n where C(n, r) =combinations of n things, taken r at a time (= Cr). Now, comparing the coeﬃcients of pn−r from right side of (1), and that from (2), we get,

S(n + 1, r + 1) = C(n + 1, r + 1)S(n, 0) + C(n, r)S(n, 1) + C(n − 1, r − 1)S(n, 2) + ... + C(n − r + 1, 1)S(n, r) + ... + S(n + 1, r). because C(n − r, 0) = 1. Now, S(n + 1, r + 1) = S(n, r + 1) + (n + 1)S(n, r), from (1.4) above. Hence,

S(n, r + 1) + (n + 1)S(n, r) = C(n + 1, r + 1)S(n, 0) + C(n, r)S(n, 1) + C(n − 1, r − 1) S(n, 2) + ... + C(n − r + 1, 1)S(n, r) + ... + S(n + 1, r), Vol. 2 Smarandache sums of products 99 or

(n + 1)S(n, r) − C(n − r + 1, 1)S(n, r) = C(n + 1, r + 1)S(n, 0) + C(n, r)S(n, 1) + ... + C(n − 1, r − 1)S(n, 2) + ... + C(n − r + 2, 2) S(n, r − 1), or

S(n, r)[(n + 1) − (n − r + 1)] = C(n + 1, r + 1)S(n, 0) + C(n, r)S(n, 1) + ... + C(n − 1, r − 1)S(n, 2) + ... + C(n − r + 2, 2)S(n, r − 1), that is

r · S(n, r) = C(n + 1, r + 1)S(n, 0) + C(n, r)S(n, 1) + C(n − 1, r − 1)S(n, 2) + ... + C(n − r + 2, 2)S(n, r − 1).

This is the required result. Veriﬁcation: Put r = 2,

S(n, 2) = C(n + 1, 3) · S(n, 0) + C(n, 2) · S(n, 1) = (n + 1)n(n − 1)/6 + n(n − 1) · n · (n + 1)/2, or S(n, 2) = (n − 1) · n · (n + 1) · (3n + 2)/24.

For n = 5, S(5, 2) = 4 · 5 · 6 · 17/24 = 85, which is true. For n = 4, S(4, 2) = 3 · 4 · 5 · 14/24 = 35, which is also true.

2.3 Reduction formula for S(n, 2). We have, S(n, r) = S(n − 1, r) + nS(n − 1, r − 1), r < n.

Put r = 2, S(n, 2) = S(n − 1, 2) + nS(n − 1, 1),

S(n − 1, 2) = S(n − 2, 2) + (n − 1)S(n − 2, 1),

S(n − 2, 2) = S(n − 3, 2) + (n − 2)S(n − 3, 1),

···

S(3, 2) = S(2, 2) + 3S(2, 1), 100 Anant W. Vyawahare No. 3

S(2, 2) = 0 + 2S(1, 1). Now, S(1, 1) = 1. Hence adding these results, Xn S(n, 2) = 2 + pS(p − 1, 1)/2 3 Xn = 2 + p(p − 1)p/2 3 Xn = 2 + (p3 − p2)/2 − (1/2)[(13 + 23) − (12 + 22)] 1 = 2 + n2(n + 1)2/8 − n(n + 1)(2n + 1)/12 − (9 − 5)/2 = (n − 1)n(n + 1)(3n + 2)/24.

This results is already obtain. 2.4 Similarly, S(n, 3) = C(n + 1, 4)3n(n + 1)/64, S(n, 4) = C(n + 1, 5)(15n3 + 15n2 − 10n + 8)/48, S(n, 5) = C(n + 1, 6)(3n4 + 2n3 − 7n2 − 6n)/16.

3.1 Definition. E(n, r) is sum of products of first n even natural numbers, taken r at a time. O(n, r) is sum of products of first n odd natural numbers, taken r at a time. We have,

(p + 2) · (p + 4) · (p + 6) = p3 + 12p2 + 44p + 48 + ... = E(3, 0)p3 + E(3, 1)p2 + E(3, 2)p + E(3, 3), (3) where, E(3, 0) = 1, E(3, 1) = 2 + 4 + 6 = 12, E(3, 2) = 2 · 4 + 2 · 6 + 4 · 6 = 44, E(3, 3) = 2 · 4 · 6 = 48. Now, put p = 1 in (3), Left side is 3 · 5 · 7 = O(3, 3).

Hence, O(3, 3) = E(3, 0) + E(3, 1) + E(3, 2) + E(3, 3) = 105,

O(5, 5) = E(4, 0) + E(4, 1) + E(4, 2) + E(4, 3) + E(4, 4) = 945. Similarly, we have,

E(3, 3) = O(3, 0) + O(3, 1) + O(3, 2) + O(3, 3) = 48.

Therefore, in general, it can be conjectured that , Vol. 2 Smarandache sums of products 101

Xn O(n + 1, n + 1) = E(n, r), r=0 and Xn E(n, n) = O(n, r). r=0 3.2 Now we extend the deﬁnition of summation for a set of natural numbers, not necessarily beginning from 1. S[(p + 1, p + n), r] is sum of products of n natural numbers, beginning from a natural number p + 1, taken r at a time. We state,

S[(p + 1, p + n), 0] + S[(p + 1, p + n), 1] + S[(p + 1, p + n), 2] + ... + S[(p + 1, p + n), r] + ... + S[(p + 1, p + n), n] = S[(p + 1, p + n + 1), r] .... (4)

Veriﬁcation: Put p = n = 3. To verify:

S[(3, 6), 0] + S[(3, 6), 1] + S[(3, 6), 2] + S[(3, 6), 3] + S[(3, 6), 4] = S[(4, 7), 4].

Left side is

1 + (3 + 4 + 5 + 6) + (3 · 4 + 3 · 5 + 3 · 6 + 4 · 5 + 4 · 6 + 5 · 6) +(3 · 4 · 5 + 3 · 4 · 6 + 4 · 5 · 6 + 4 · 5 · 6) + (3 · 4 · 5 · 6) = 1 + 18 + 119 + 342 + 360 = 840.

Right side is 4 · 5 · 6 · 7 = 840. Hence veriﬁed. Hence the result (4) is true. Similar results are true for odd and even integers,as

O[(2, 7), 0] + O[(2, 7), 1] + O[(2, 7), 2] + O[(2, 7), 3] + O[(2, 7), 4] = E[(3, 8), 4],

and E[(2, 7), 0] + E[(2, 7), 1] + E[(2, 7), 2] + E[(2, 7), 3] + E[(2, 7), 4] = O[(3, 8), 4]. The veriﬁcation of these results are simple. 3.3 To prove: 1. S(n − 1, 1) = nS(n, 1) − n, 2. S(n − 1, 2) = S(n, 2) − nS(n, 1) + n2, 3. S(n − 1, 3) = S(n, 3) − nS(n, 2) + n2S(n, 1) − n3, 102 Anant W. Vyawahare No. 3

··· 4. S(n − 1, r) = S(n, r) − nS(n, r − 1) + n2S(n, r − 2) − n3S(n, r − 3) + ... Proof. We have

(p + 1)(p + 2)(p + 3) ... (p + n − 1)(p + n) = S(n, 0)pn + S(n, 1)pn−1 + S(n, 2)pn−2 + S(n, 3)pn−3 + ..., (5)

divide both sides by (p + n),

(p + 1)(p + 2)(p + 3) ... (p + n − 1) = [S(n, 0)pn + S(n, 1)pn−1 + S(n, 2)pn−2 + S(n, 3)pn−3 + ...]/(p + n). (6)

Now using result (5), the left side of (6) is

S(n − 1, 0)pn−1 + S(n − 1, 1)pn−2 + S(n − 1, 2)pn−3 + S(n − 1, 3)pn−4 + .... (7)

By actual division, the right side of (6) is

pn−1 + pn−2 · [S(n, 1) − n] + pn−3[S(n, 2) − n · S(n, 1) + n] + .... (8)

Equating the coeﬃcients of like powers of p from (7) and (8), we have, S(n − 1, 1) = n · S(n, 1) − n, S(n − 1, 2) = S(n, 2) − n · S(n, 1) + n S(n − 1, 3) = S(n, 3) − n · S(n, 2) + n2 · S(n, 1) − n3 ··· . S(n − 1, r) = S(n, r) − n · S(n, r − 1) + n2S(n, r − 2) − n3 · S(n, r − 3) + ··· . Veriﬁcation : For n = 5, and r = 3, right side is S(4, 3) = 50, also, left side is

S(5, 3) − 5 · S(5, 2) + 52 · S(5, 1) − 53 = 225 − 5(85) + 25(15) = 50.

We have, S(n, r) = S(n − 1, r) + n · S(n − 1, r − 1), r < n. Also S(n − 1, r) = S(n − 2, r) + (n − 1) · S(n − 2, r − 1). Adding, S(n, r) = S(n − 2, r) + (n − 1) · S(n − 2, r − 1) + n · S(n − 1, r − 1). Again, S(n − 1, r − 1) = S(n − 2, r − 1) + n − 1 · S(n − 2, r − 2), hence,

S(n, r) = S(n − 2, r) + n − 1 · S(n − 2, r − 1) + n · S(n − 2, r − 1) + n · (n − 1)S(n − 2, r − 2), Vol. 2 Smarandache sums of products 103

S(n, r) = S(n − 2, r) + (2n − 1) · S(n − 2, r − 1) + n · (n − 1) · S(n − 2, r − 2).

Veriﬁcation: Put n = 5 and r = 3, the left side is

S(5, 3) = 225, right side is S(3, 3) + (9) · S(3, 2) + 5 · 4 · S(3, 1) = 6 + 9.11 + 5.4.6 = 225, hence veriﬁed. 4.1 Interestingly, the set of numbers S(n, r), forms a sequence when r is ﬁxed. For, S(2, 2) = 2,S(3, 2) = 11,S(4, 2) = 35,S(5, 2) = 85,S(6, 2) = 175 ... The numbers

{2, 11, 35, 85, 175, 322, 546, 870, 1320, 1925, 2717, 3731,..., } are the Stirling numbers of ﬁrst kind. These numbers are the numbers of edges of a complete k-partite graph of order S(k, 1), that is of order k(k + 1)/2. The n-th term of this sequence is given by

an = an−1 + [n(n + 1)2]/2, n ≥ 2, and a1 = 2. 4.2 A similar sequence for S(n, 3) is

{6, 50, 225, 1960, 4536, 9450, 18150, 32670,...} with similar properties , thus the sequences generated by S(n, r) create additional good results. These are also Stirling numbers.

References

[1] S.H.Ramsubramanian, Patterns in product summations, AMTI, Chennai, India, 1991. Scientia Magna Vol. 2 (2006), No. 3, 104-108

On the solutions of an equation involving the Smarandache dual function

Na Yuan

Department of Mathematics, Northwest University Xi’an, Shaanxi, P.R.China

Abstract In this paper, we use the elementary method to study the positive integer solutions

of an equation involving the Smarandache dual functions ¯k(n), and give its all solutions. Keywords Smarandache dual function, the positive integer solutions.

§1. Introduction

For any positive integer n, the famous Smarandache function S(n) is deﬁned by

S(n) = min{m ∈ N : n | m!}.

For example, S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 4, S(5) = 5, S(6) = 3, S(7) = 7, S(8) = 4, ··· . This function was introduced by American-Romanian number theorist professor F.Smarandache, see reference [1]. About the arithmetical properties of S(n), many scholars had studied it, and obtained some interesting conclusions, see references [2] and [3]. At the same time, many scholars also studied another function which have close relations with the Smarandache function. It is called the Smarandache ceil function Sk(n), we deﬁne this arithmetic function as follows:

k sk(n) = min{m ∈ N : n | m }.

For example, if k = 3, we have the sequence {s3(n)} (n = 1, 2, 3, ··· ) as following: s3(1) =

1, s3(2) = 2, s3(3) = 3, s3(4) = 2, s3(5) = 5, s3(6) = 6, s3(7) = 7, s3(8) = 2, ··· . This arithmetical function is a multiplicative function, and has many interesting properties, so it had be studied by many people, see references [4] and [5].

Similarly, we will introduce the Smarandache dual functions ¯k(n) which denotes the greatest positive integer m such that mk|n, where n denotes any positive integer. That is,

k s¯k(n) = max{m ∈ N : m |n}.

It is easy to calculating thats ¯3(1) = 1,s ¯3(2) = 1,s ¯3(3) = 1,s ¯3(4) = 1,s ¯3(5) = 1, s¯3(6) = 1, s¯3(7) = 1,s ¯3(8) = 2,X··· . About this function, Lu Yaming [6] studied the asymptotic properties of the summation d(¯sk(n)) by using the elementary methods, and obtained an n≤x Vol. 2 On the solutions of an equation involving the Smarandache dual function 105

interesting asymptotic formula: µ ¶ ³ ´ X 1 1 1 d (¯s (n)) = ζ(k)x + ζ x k + O x k+1 . k k n≤x X Ding Liping [7] also studied the mean value properties of the summation s¯2(a(n)), and give n≤x a sharper mean value theorem: µ ¶ X 2 4 Y ³ ´ x π 1 3 +ε s¯ (a(n)) = 1 + + O x 2 , 2 315 p4 + p3 n≤x p where a(n) denotes the cubic complements of n, ζ(s) is the Riemann zeta-function. On the other hand, we let Ω(n) denotes the number of the prime divisors of n, including

α1 α2 αr multiple numbers. If n = p1 p2 ··· pr denotes the factorization of n into prime powers, then

Ω(n) = α1 + α2 ··· + αr, we have Ω(1) = 0, Ω(2) = 1, Ω(3) = 1, Ω(4) = 2, Ω(5) = 1, Ω(6) = 2, Ω(7) = 1, Ω(8) = 3, ··· . In [8], the author studied the positive integer solutions of the equation

s¯3(1) +s ¯3(2) + ··· +s ¯3(n) = 3Ω(n),

and proved that this equation has three solutions n = 3, 6 and 8. This paper, as a note of [8], we shall prove a general conclusion for the equation

s¯k(1) +s ¯k(2) + ··· +s ¯k(n) = kΩ(n).

That is, we shall prove the following: Theorem. For all positive integer n, the equation

s¯k(1) +s ¯k(2) + ··· +s ¯k(n) = kΩ(n)

has no solution if k = 1. If k ≥ 2, then the equation has at least one positive integer solution. They are i) n = 2, if k = 2; ii) n = 3, 6, 8, if k = 3; iii) If k ≥ 4 and n < 2k, then the equation has solution if and only if n = kΩ(n). Especially, if k = p be any prime, then n = p and 2p are two solutions of the equation.

§2. Some lemmas

To complete the proof of the theorem, we need the following lemmas: Lemma 1. For all positive integer n, the equation

s¯1(1) +s ¯1(2) + ··· +s ¯1(n) = Ω(n), 106 Na Yuan No. 3

has no positive integer solution. Proof. For any positive integer n, if k = 1, then we have

s1(n) = n, Xn and then s1(i) ≥ n. i=1 If n = 1, thens ¯1(1) = 1 and Ω(1) = 0, this time we know that the equation has no positive α1 α2 αr integer solution; If n ≥ 2 n = p1 p2 ··· pr and pi is a prime, then

α1 α2 αr n = p1 p2 ··· pr > α1 · α2 ···· αr.

Note that if a > 1, b > 1, then a · b ≥ a + b, so we have Xn s1(i) > α1 · α2 ···· αr > α1 + α2 + ··· + αr = Ω(n). i=1

So for all positive integer n, from the above formula we know that the equations ¯1(1) +s ¯1(2) +

··· +s ¯1(n) = Ω(n) has no positive integer solution. This proved Lemma 1. Lemma 2. For all positive integer n, the equation

s¯2(1) +s ¯2(2) + ··· +s ¯2(n) = 2Ω(n) has one solution n = 2; The equation

s¯3(1) +s ¯3(2) + ··· +s ¯3(n) = 3Ω(n) has three solutions, n = 3, 6, 8. Proof. The second result can be deduced directly from reference [8]. By the same way we can also get the ﬁrst result of Lemma 2. Lemma 3. For all integers k ≥ 4, we have 2k+1 > k(k + 1). Proof. Now we prove this inequality by mathematical induction on k. For k = 4, it holds trivially. Suppose, then, that the formula holds for all integers < k. That is, for all integers m < k, we have 2m+1 > m(m + 1). Note that 2m(m + 1) > (m + 1)(m + 2), from the inductive hypothesis we have

2m+1+1 > 2m(m + 1) > (m + 1)(m + 2).

That is means, the inequality also holds for m + 1. This proved Lemma 3. Lemma 4. If k ≥ 4, then for all positive integer n ≥ 2k, we have

s¯k(1) +s ¯k(2) + ··· +s ¯k(n) > kΩ(n).

α1 α2 αr Proof. Let n = p1 p2 ··· pr be the factorization of n into prime powers, then we have

k s¯k(1) +s ¯k(2) + ··· +s ¯k(n) > n if n ≥ 2 . Vol. 2 On the solutions of an equation involving the Smarandache dual function 107

From the deﬁnition of Ω(n), we have

Ω(n) = α1 + α2 ··· + αr.

So to complete the proof of the lemma, we only to prove the following inequality:

α1 α2 αr p1 p2 ··· pr > k(α1 + α2 ··· + αr). (1)

Now we prove (1) by mathematical induction on r.

α1 i) If r = 1, then n = p1 . k+1 k k+1 a. If p1 = 2, since 2 > 2 , then we have α1 ≥ k+1, and Lemma 3 tell us 2 > k(k+1). Hence

α1 2 > kα1.

k k 4 2 b. If p1 ≥ 3, then p1 > 2 . So we have α1 ≥ k. If k = 4, we have p > 4 , and if k > 4, we have pk > 4k > k2. So we know k p1 > k · k.

Hence

α1 p1 > kα1.

This proved that the lemma 4 holds for r = 1. ii) Now we assume that the (1) holds for r (≥ 2), we shall prove that it is also holds for r + 1. From the inductive hypothesis, we have

α1 α2 αr αr+1 αr+1 p1 p2 ··· pr pr+1 > k(α1 + α2 + ··· + αr) · pr+1 .

Since pr+1 is a prime, then

αr+1 pr+1 > αr+1 + 1.

From above we obtain

α1 α2 αr αr+1 p1 p2 ··· pr pr+1 > k(α1 + α2 + ··· + αr) · (αr+1 + 1).

Note that if a > 1, b > 1, then a · b ≥ a + b, so we have

(α1 + α2 ··· + αr) · (αr+1 + 1) ≥ α1 + α2 + ··· + αr + αr+1 + 1 > α1 + α2 + ··· + αr + αr+1.

α1 α2 αr αr+1 p1 p2 ··· pr pr+1 > k(α1 + α2 + ··· + αr + αr+1).

This completes the proof of Lemma 4. 108 Na Yuan No. 3 §3. Proof of the theorem

In this section, we shall complete the proof of the theorem. According to the deﬁnition of s¯k(n) and the results of Lemma 1 and Lemma 2, we only to prove the following case: when k ≥ 4, whether there exists ﬁnite solutions for the equation

s¯k(1) +s ¯k(2) + ··· +s ¯k(n) = kΩ(n).

First we separate all positive integer into two cases: k 1. If n < 2 , then from the deﬁnition ofs ¯k(n) and Ω(n), we haves ¯k(n) = 1, so the equation α1 α2 αr s¯k(1) +s ¯k(2) + ··· +s ¯k(n) = kΩ(n) become the form n = kΩ(n). Hence if n = p1 p2 ··· pr , then n = k(α1 + α2 + ··· + αr) are the positive integer solutions of the equation. 2. If n ≥ 2k, then from Lemma 4 we know that the equation has no positive integer solutions. Combining all the above cases we have the following conclusion: The equation

s¯k(1) +s ¯k(2) + ··· +s ¯k(n) = kΩ(n) has no solution if k = 1. If k ≥ 2, then the equation has positive integer solutions. They are i) n = 2, if k = 2; ii) n = 3, 6, 8, if k = 3; iii) If k ≥ 4 and n < 2k, then the equation has solution if and only if n = kΩ(n). Note: It is clear that if k = p be a prime, then n = p or 2p are two solutions of the equation. This completes the proof of Theorem.

References

[1] F. Smarandache, Only problems, not Solutions, Xiquan Publ. House, Chicago, 1993. [2] Li Hailong and Zhao Xiaopeng, On the Smarandache function and the k-th roots of a positive integer, Research on Smarandache Problems in Number Theory. Hexis, 2004, 119-122. [3] Mark Farris and Patrick Mitchell, Bounding the Smarandache function, Smarandache Notions Journal, Vol. 13(2002), 37-42. [4] Sadin Tabirce and Tatiana Tairca, Some new results concerning the Smarandache ceil function, Smarandache Notions Journal, Vol. 13(2002), 30-36. [5] Ren Dongmei, On the Smarandache ceil function and the Dirichlet divisor function, Smarandache Problems in Number Theory, Hexis, (2005), 51-54. [6] Lu Yaming, On a dual function of the Smarandache ceil function, Smarandache Prob- lems in Number Theory, Hexis, (2005), 55-57. [7] Ding Liping, An arithmetical function and cubic complements, Smarandache Problems in Number Theory, Hexis, (2004), 93-95. [8] Yuan Na, On the solutions of an equation involving the Smarandache dual function, Scientia Magna, Vol. 2(2006), No. 2, 27-30. Scientia Magna Vol. 2 (2006), No. 3, 109-112

An inﬁnite series involving the Smarandache power function SP (n)

Huanqin Zhou

Department of Mathematics, Weinan Teacher’s College Weinan, Shaanxi, P.R.China

Abstract For any positive integer n, the Smarandache power function SP (n) is defined as the smallest positive integer m such that n|mm, where m and n have the same prime divisors. The main purpose of this paper is using the elementary methods to study the convergent properties of an infinite series involving the Smarandache power function SP (n), and give some interesting identities. Keywords Smarandache power function, infinite series, the Riemann zeta-function.

§1. Introduction and Results

For any positive integer n, we deﬁne the Smarandache power function SP (n) as the smallest positive integer m such that n|mm, where n and m have the same prime divisors. That is,    Y Y  SP (n) = min m : n|mm, m ∈ N, p = p .   p|n p|m

If n runs through all natural numbers, then we can get the Smarandache power function sequence {SP (n)}: 1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, ··· . In reference [1], Professor F.Smarandache asked us to study the properties of the sequence {SP (n)}. From the deﬁnition of SP (n) we can easily get the following conclusions: If n = pα, where p be a prime, then we have    p, if 1 ≤ α ≤ p;   2 2  p , if p + 1 ≤ α ≤ 2p ; SP (n) = p3, if 2p2 + 1 ≤ α ≤ 3p3;    ···   pα, if (α − 1)pα + 1 ≤ α ≤ αpα.

α1 α2 αr Let n = p1 p2 ··· pr denotes the factorization of n into primeY powers.Y If αi ≤ pi for all αi (i = 1, 2, ··· , r), then we have SP (n) = U(n), where U(n) = p, denotes the product p|n p|n over all diﬀerent prime divisors of n. It is clear that SP (n) is not a multiplicative function. For example, SP (8) = 4, SP (3) = 3, SP (24) = 6 6= SP (3) × SP (8). But for almost m and n with 110 Huanqin Zhou No. 3

(m, n) = 1, we have SP (mn) = SP (m) · SP (n). In reference [2], Dr. Zhefeng Xu had studied the mean value properties of SP (n), and obtained some sharper asymptotic formulas, one of them as follows:

X Y µ ¶ ³ ´ 1 2 1 3 +² SP (n) = x 1 − + O x 2 , 2 p(p + 1) n≤x p Y where ² denotes any fixed positive number, and denotes the product over all primes. p In this paper, we shall use the elementary methods to study the convergent properties of an infinite series involving the Smarandache power function SP (n), and give some interesting identities. That is, we shall prove the following: Theorem. For any complex number s with Re(s) > 1, we have the identities  2s+1 1  s , if k = 1, 2; X∞ n−1  2 −1 ζ(s) (−1) µ(n) 2s+1 1 2s−1 s = s − s , if k = 3; (SP (nk))  2 −1 ζ(s) 4 n=1  2s+1 1 2s−1 3s−1 2s−1 ζ(s) − 4s + 9s , if k = 4, 5. where µ(n) denotes the Möbius function, and ζ(s) denotes the Riemann zeta-function. π2 π4 Note that ζ(2) = 6 , ζ(4) = 90 , taking s = 2 and 4 in our Theorem we may immediately deduce the following identities:  10  2 , if k = 1, 2; X∞ n−1  π (−1) µ(n) 10 3 2 = 2 − , if k = 3; (SP (nk))  π 16 n=1  10 3 8 π2 − 16 + 81 , if k = 4, 5. and  102  4 , if k = 1, 2; X∞ n−1  π (−1) µ(n) 102 15 4 = 4 − , if k = 3; (SP (nk))  π 256 n=1  102 15 80 π4 − 256 + 6561 , if k = 4, 5. Note: For general integer k ≥ 6, using our method we can also give a calculating formula X∞ (−1)n−1µ(n) for , but in this case, the conclusion is more complicate. (SP (nk))s n=1

§2. Proof of the theorem

In this section, we shall prove our Theorem directly. Note that µ(2m) = 0 if m be an even number. If m be a odd number, then µ(2m) = −µ(m). So we have

X∞ (−1)n−1µ(n) X∞ µ(2m − 1) X∞ µ(2m) = − (SP (nk))s (SP ((2m − 1)k))s (SP ((2m)k))s n=1 m=1 m=1 X∞ µ(2m − 1) X∞ µ(2m − 1) = + . (1) (SP ((2m − 1)k))s (SP (2k(2m − 1)k))s m=1 m=1 Vol. 2 An inﬁnite series involving the Smarandache power function SP (n) 111

µ(n) If k = 1 or 2, then is a multiplicative function. In fact for any positive integers m SP (nk) and n with (m, n) = 1, if µ(mn) = 0, then µ(m) = 0 or µ(n) = 0, i.e. µ(m)µ(n) = 0. So

µ(mn) µ(m) µ(n) µ(m) µ(n) = = . SP (mknk) SP (mk) SP (nk) m n

If µ(mn) 6= 0, then µ(m) 6= 0, µ(n) 6= 0, and more SP (mknk) = SP (mk)SP (nk) = mn. Thus, µ(n) µ(n) = is a multiplicative function, if k = 1 or 2. SP (nk) n Now if k = 1 or 2, note that the identity (see Theorem 11.7 of [3]) µ ¶ X∞ µ(n) Y 1 1 = 1 − = , ns ps ζ(s) n=1 p

where ζ(s) is the Riemann zeta-function, then from (1), the Euler product formula (see Theorem µ(n) 11.6 of [3]) and the multiplicative properties of we have (SP (nk))s

X∞ (−1)n−1µ(n) X∞ µ(2m − 1) X∞ µ(2m − 1) = + (SP (nk))s (SP ((2m − 1)k))s (SP (2k(2m − 1)k))s n=1 m=1µ ¶ µ m=1 ¶ Y 1 1 Y 1 = 1 − + 1 − ps 2s ps p6=2 p6=2 µ ¶ 2s + 1 2s Y 1 2s + 1 X∞ µ(n) = 1 − = 2s 2s − 1 ps 2s − 1 ns p n=1 2s + 1 1 = . 2s − 1 ζ(s)

This proves the ﬁrst formula of our Theorem. µ(1) 1 µ(2m − 1) µ(2m − 1) If k = 3, note that SP (23) = 4, so = , and = (SP (23))s 4s (SP (23(2m − 1)3))s 2s(2m − 1)s for all m > 1. Then from (1) and the method of proving the ﬁrst part of our Theorem we have

X∞ (−1)n−1µ(n) X∞ µ(2m − 1) X∞ µ(2m − 1) = + (SP (n3))s (SP ((2m − 1)3))s (SP (23(2m − 1)3))s n=1 m=1 m=1 X∞ µ(2m − 1) 1 1 X∞ µ(2m − 1) = + − + (SP ((2m − 1)3))s 4s 2s 2s(2m − 1)s m=1µ ¶ µ m=1 ¶ Y 1 2s − 1 1 Y 1 = 1 − − + 1 − ps 4s 2s ps p6=2 p6=2 µ ¶ 2s + 1 2s Y 1 2s − 1 = 1 − − 2s 2s − 1 ps 4s p 2s + 1 1 2s − 1 = − . 2s − 1 ζ(s) 4s

This proves the second formula of Theorem. Now we prove the third formula of our Theorem. 112 Huanqin Zhou No. 3

µ(1) 1 µ(3) For k = 4 or 5, note that SP (2k) = 4, SP (3k) = 9, i.e., = , = (SP (2k))s 4s (SP (3k))s 1 µ(2m − 1) µ(2m − 1) µ(2m − 1) µ(2m − 1) − , and = for all m > 2, = 9s (SP (2k(2m − 1)k))s 2s(2m − 1)s (SP ((2m − 1)k))s (2m − 1)s for all m ≥ 2. So from the Euler product formula and (1) we have

X∞ (−1)n−1µ(n) X∞ µ(2m − 1) X∞ µ(2m − 1) = + (SP (nk))s (SP ((2m − 1)k))s (SP (2k(2m − 1)k))s n=1 m=1 m=1 1 1 X∞ µ(2m − 1) 1 1 X∞ µ(2m − 1) = − + + + − + 9s 3s (2m − 1)s 4s 2s 2s(2m − 1)s m=1µ ¶ m=1µ ¶ 1 1 Y 1 2s − 1 1 Y 1 = − + + 1 − − + 1 − 9s 3s ps 4s 2s ps p6=2 p6=2 µ ¶ 2s + 1 2s Y 1 2s − 1 3s − 1 = 1 − − + 2s 2s − 1 ps 4s 9s p 2s + 1 1 2s − 1 3s − 1 = − + . 2s − 1 ζ(s) 4s 9s

This completes the proof of our Theorem.

References

[1] F.Smarandache, Collected papers, Bucharest, Tempus Publ. Hse., 1998. [2] Zhefeng Xu, On the mean value of the Smarandache power function, Acta Mathematica Sinica (Chinese series), 49(2006), No. 1, pp. 77-80. [3] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976. [4] Chengdong Pan and Chengbiao Pan, The Elementary Number Theory, Beijing Univer- sity Press, Beijing, 2003.

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